The basic properties of the natural logarithm, graph, domain of definition, set of values, basic formulas, derivative, integral, expansion in power series and representation of the function ln x using complex numbers.
Definition
Natural logarithm is the function y = ln x, inverse to exponential, x = e y , and is logarithm based on the number e: ln x = log e x.
The natural logarithm is widely used in mathematics because its derivative has the simplest form: (ln x)′ = 1/ x.
Based definitions, the base of the natural logarithm is the number e:
e ≅ 2.718281828459045...;
.
Graph of the function y = ln x.
Graph of natural logarithm (functions y = ln x) is obtained from exponential graphics mirror image relative to the straight line y = x.
The natural logarithm is defined at positive values variable x. It increases monotonically in its domain of definition.
At x → 0 the limit of the natural logarithm is minus infinity (-∞).
As x → + ∞, the limit of the natural logarithm is plus infinity (+ ∞). For large x, the logarithm increases quite slowly. Any power function x a s positive indicator degree a grows faster than the logarithm.
Properties of the natural logarithm
Domain of definition, set of values, extrema, increase, decrease
The natural logarithm is a monotonically increasing function, so it has no extrema. The main properties of the natural logarithm are presented in the table.
ln x values
ln 1 = 0
Basic formulas for natural logarithms
Formulas following from the definition of the inverse function:
The main property of logarithms and its consequences
Base replacement formula
Any logarithm can be expressed in terms of natural logarithms using the base replacement formula:
Proofs of these formulas are presented in section "Logarithm".
Inverse function
The inverse of the natural logarithm is exponent.
If , then
If, then.
Derivative ln x
Derivative of the natural logarithm:
.
Derivative of the natural logarithm of modulus x:
.
Derivative of nth order:
.
Deriving formulas > > >
Integral
The integral is calculated integration by parts :
.
So,
Expressions using complex numbers
Consider the function of the complex variable z:
.
Let's express the complex variable z via module r and argument φ
:
.
Using the properties of the logarithm, we have:
.
Or
.
The argument φ is not uniquely defined. If you put
, where n is an integer,
it will be the same number for different n.
Therefore, the natural logarithm, as a function of a complex variable, is not a single-valued function.
Power series expansion
When the expansion takes place:
References:
I.N. Bronstein, K.A. Semendyaev, Handbook of mathematics for engineers and college students, “Lan”, 2009.
We continue to study logarithms. In this article we will talk about calculating logarithms, this process is called logarithm. First we will understand the calculation of logarithms by definition. Next, let's look at how the values of logarithms are found using their properties. After this, we will focus on calculating logarithms through initially set values other logarithms. Finally, let's learn how to use logarithm tables. The entire theory is provided with examples with detailed solutions.
Page navigation.
Calculating logarithms by definition
In the simplest cases it is possible to perform quite quickly and easily finding the logarithm by definition. Let's take a closer look at how this process happens.
Its essence is to represent the number b in the form a c, from which, by the definition of a logarithm, the number c is the value of the logarithm. That is, by definition, the following chain of equalities corresponds to finding the logarithm: log a b=log a a c =c.
So, calculating a logarithm by definition comes down to finding a number c such that a c = b, and the number c itself is the desired value of the logarithm.
Taking into account the information in the previous paragraphs, when the number under the logarithm sign is given by a certain power of the logarithm base, you can immediately indicate what the logarithm is equal to - it equal to the indicator degrees. Let's show solutions to examples.
Example.
Find log 2 2 −3, and also calculate the natural logarithm of the number e 5,3.
Solution.
The definition of the logarithm allows us to immediately say that log 2 2 −3 =−3. Indeed, the number under the logarithm sign is equal to base 2 to the −3 power.
Similarly, we find the second logarithm: lne 5.3 =5.3.
Answer:
log 2 2 −3 =−3 and lne 5,3 =5,3.
If the number b under the logarithm sign is not specified as a power of the base of the logarithm, then you need to carefully look to see if it is possible to come up with a representation of the number b in the form a c . Often this representation is quite obvious, especially when the number under the logarithm sign is equal to the base to the power of 1, or 2, or 3, ...
Example.
Calculate the logarithms log 5 25 , and .
Solution.
It is easy to see that 25=5 2, this allows you to calculate the first logarithm: log 5 25=log 5 5 2 =2.
Let's move on to calculating the second logarithm. The number can be represented as a power of 7: 
(see if necessary). Hence, 
.
Let's rewrite the third logarithm in the following form. Now you can see that 
, from which we conclude that 
. Therefore, by the definition of logarithm 
.
Briefly, the solution could be written as follows: .
Answer:
log 5 25=2 , 
And .
When under the logarithm sign there is a sufficiently large natural number, then it wouldn’t hurt to decompose it into prime factors. It often helps to represent such a number as some power of the base of the logarithm, and therefore calculate this logarithm by definition.
Example.
Find the value of the logarithm.
Solution.
Some properties of logarithms allow you to immediately specify the value of logarithms. These properties include the property of the logarithm of a unit and the property of the logarithm of a number, equal to the base: log 1 1=log a a 0 =0 and log a a=log a a 1 =1 . That is, when under the sign of the logarithm there is a number 1 or a number a equal to the base of the logarithm, then in these cases the logarithms are equal to 0 and 1, respectively.
Example.
What are logarithms and log10 equal to?
Solution.
Since , then from the definition of logarithm it follows 
.
In the second example, the number 10 under the logarithm sign coincides with its base, so the decimal logarithm of ten equal to one, that is, log10=lg10 1 =1.
Answer:
AND lg10=1 .
Note that the calculation of logarithms by definition (which we discussed in previous paragraph) implies the use of the equality log a a p =p, which is one of the properties of logarithms.
In practice, when a number under the logarithm sign and the base of the logarithm are easily represented as a power of a certain number, it is very convenient to use the formula 
, which corresponds to one of the properties of logarithms. Let's look at an example of finding a logarithm that illustrates the use of this formula.
Example.
Calculate the logarithm.
Solution.
Answer:
.
Properties of logarithms not mentioned above are also used in calculations, but we will talk about this in the following paragraphs.
Finding logarithms through other known logarithms
The information in this paragraph continues the topic of using the properties of logarithms when calculating them. But here the main difference is that the properties of logarithms are used to express the original logarithm in terms of another logarithm, the value of which is known. Let's give an example for clarification. Let's say we know that log 2 3≈1.584963, then we can find, for example, log 2 6 by doing a little transformation using the properties of the logarithm: log 2 6=log 2 (2 3)=log 2 2+log 2 3≈ 1+1,584963=2,584963 .
In the above example, it was enough for us to use the property of the logarithm of a product. However, much more often it is necessary to use a wider arsenal of properties of logarithms in order to calculate the original logarithm through the given ones.
Example.
Calculate the logarithm of 27 to base 60 if you know that log 60 2=a and log 60 5=b.
Solution.
So we need to find log 60 27 . It is easy to see that 27 = 3 3 , and the original logarithm, due to the property of the logarithm of the power, can be rewritten as 3·log 60 3 .
Now let's see how to express log 60 3 in terms of known logarithms. The property of the logarithm of a number equal to the base allows us to write the equality log 60 60=1. On the other hand, log 60 60=log60(2 2 3 5)= log 60 2 2 +log 60 3+log 60 5= 2·log 60 2+log 60 3+log 60 5 . Thus, 2 log 60 2+log 60 3+log 60 5=1. Hence, log 60 3=1−2·log 60 2−log 60 5=1−2·a−b.
Finally, we calculate the original logarithm: log 60 27=3 log 60 3= 3·(1−2·a−b)=3−6·a−3·b.
Answer:
log 60 27=3·(1−2·a−b)=3−6·a−3·b.
Separately, it is worth mentioning the meaning of the formula for transition to a new base of the logarithm of the form 
. It allows you to move from logarithms with any base to logarithms with a specific base, the values of which are known or it is possible to find them. Usually, from the original logarithm, using the transition formula, they move to logarithms in one of the bases 2, e or 10, since for these bases there are tables of logarithms that allow their values to be calculated with a certain degree of accuracy. IN next point we'll show you how it's done.
Logarithm tables and their uses
For approximate calculation of logarithm values can be used logarithm tables. The most commonly used base 2 logarithm table, natural logarithm table, and decimal logarithms. When working in decimal system For calculus, it is convenient to use a table of logarithms based on base ten. With its help we will learn to find the values of logarithms.
The presented table allows you to find the values of the decimal logarithms of numbers from 1,000 to 9,999 (with three decimal places) with an accuracy of one ten-thousandth. We will analyze the principle of finding the value of a logarithm using a table of decimal logarithms into specific example– it’s clearer that way. Let's find log1.256.
In the left column of the table of decimal logarithms we find the first two digits of the number 1.256, that is, we find 1.2 (this number is circled in blue for clarity). We find the third digit of 1.256 (digit 5) in the first or last line to the left of the double line (this number is circled in red). The fourth digit of the original number 1.256 (digit 6) is found in the first or last line to the right of the double line (this number is circled with a green line). Now we find the numbers in the cells of the table of logarithms at the intersection of the marked row and marked columns (these numbers are highlighted orange). The sum of the marked numbers gives the desired value of the decimal logarithm accurate to the fourth decimal place, that is, log1.236≈0.0969+0.0021=0.0990.
Is it possible, using the table above, to find the values of decimal logarithms of numbers that have more than three digits after the decimal point, as well as those that go beyond the range from 1 to 9.999? Yes, you can. Let's show how this is done with an example.
Let's calculate lg102.76332. First you need to write down number in standard form : 102.76332=1.0276332·10 2. After this, the mantissa should be rounded to the third decimal place, we have 1.0276332 10 2 ≈1.028 10 2, while the original decimal logarithm is approximately equal to the logarithm the resulting number, that is, we take log102.76332≈lg1.028·10 2. Now we apply the properties of the logarithm: lg1.028·10 2 =lg1.028+lg10 2 =lg1.028+2. Finally, we find the value of the logarithm lg1.028 from the table of decimal logarithms lg1.028≈0.0086+0.0034=0.012. As a result, the entire process of calculating the logarithm looks like this: log102.76332=log1.0276332 10 2 ≈lg1.028 10 2 = log1.028+lg10 2 =log1.028+2≈0.012+2=2.012.
In conclusion, it is worth noting that using a table of decimal logarithms you can calculate the approximate value of any logarithm. To do this, it is enough to use the transition formula to go to decimal logarithms, find their values in the table, and perform the remaining calculations.
For example, let's calculate log 2 3 . According to the formula for transition to a new base of the logarithm, we have . From the table of decimal logarithms we find log3≈0.4771 and log2≈0.3010. Thus, 
.
Bibliography.
- Kolmogorov A.N., Abramov A.M., Dudnitsyn Yu.P. and others. Algebra and the beginnings of analysis: Textbook for grades 10 - 11 of general education institutions.
- Gusev V.A., Mordkovich A.G. Mathematics (a manual for those entering technical schools).
Instructions
Write down the given logarithmic expression. If the expression uses the logarithm of 10, then its notation is shortened and looks like this: lg b is the decimal logarithm. If the logarithm has the number e as its base, then write the expression: ln b – natural logarithm. It is understood that the result of any is the power to which the base number must be raised to obtain the number b.
When finding the sum of two functions, you simply need to differentiate them one by one and add the results: (u+v)" = u"+v";
When finding the derivative of the product of two functions, it is necessary to multiply the derivative of the first function by the second and add the derivative of the second function multiplied by the first function: (u*v)" = u"*v+v"*u;
In order to find the derivative of the quotient of two functions, it is necessary to subtract from the product of the derivative of the dividend multiplied by the divisor function the product of the derivative of the divisor multiplied by the function of the dividend, and divide all this by the divisor function squared. (u/v)" = (u"*v-v"*u)/v^2;
If given complex function, then it is necessary to multiply the derivative of internal function and the derivative of the external one. Let y=u(v(x)), then y"(x)=y"(u)*v"(x).
Using the results obtained above, you can differentiate almost any function. So let's look at a few examples:
y=x^4, y"=4*x^(4-1)=4*x^3;
y=2*x^3*(e^x-x^2+6), y"=2*(3*x^2*(e^x-x^2+6)+x^3*(e^x-2 *x));
There are also problems involving calculating the derivative at a point. Let the function y=e^(x^2+6x+5) be given, you need to find the value of the function at the point x=1.
1) Find the derivative of the function: y"=e^(x^2-6x+5)*(2*x +6).
2) Calculate the value of the function in given point y"(1)=8*e^0=8
Video on the topic
Learn the table of elementary derivatives. This will significantly save time.
Sources:
- derivative of a constant
So, what is the difference between rational equation from the rational? If the unknown variable is under the sign square root, then the equation is considered irrational.
Instructions
The main method for solving such equations is the method of constructing both sides equations into a square. However. this is natural, the first thing you need to do is get rid of the sign. This method is not technically difficult, but sometimes it can lead to trouble. For example, the equation is v(2x-5)=v(4x-7). By squaring both sides you get 2x-5=4x-7. Solving such an equation is not difficult; x=1. But the number 1 will not be given equations. Why? Substitute one into the equation instead of the value of x. And the right and left sides will contain expressions that do not make sense, that is. This value is not valid for a square root. Therefore 1 is an extraneous root, and therefore given equation has no roots.
So, irrational equation is solved using the method of squaring both its parts. And having solved the equation, it is necessary to cut off extraneous roots. To do this, substitute the found roots into the original equation.
Consider another one.
2х+vх-3=0
Of course, this equation can be solved using the same equation as the previous one. Move Compounds equations, which do not have a square root, to the right side and then use the squaring method. solve the resulting rational equation and roots. But also another, more elegant one. Enter a new variable; vх=y. Accordingly, you will receive an equation of the form 2y2+y-3=0. That is, the usual quadratic equation. Find its roots; y1=1 and y2=-3/2. Next, solve two equations vх=1; vх=-3/2. The second equation has no roots; from the first we find that x=1. Don't forget to check the roots.
Solving identities is quite simple. To do this you need to do identity transformations until the goal is achieved. Thus, with the help of the simplest arithmetic operations the task at hand will be solved.
You will need
- - paper;
- - pen.
Instructions
The simplest of such transformations are algebraic abbreviated multiplications (such as the square of the sum (difference), difference of squares, sum (difference), cube of the sum (difference)). In addition, there are many and trigonometric formulas, which are essentially the same identities.
Indeed, the square of the sum of two terms equal to square the first plus double the product of the first by the second and plus the square of the second, that is (a+b)^2= (a+b)(a+b)=a^2+ab +ba+b^2=a^2+2ab +b^2.
Simplify both
General principles of the solution
Repeat according to the textbook mathematical analysis or higher mathematics, which is a definite integral. As is known, the solution definite integral there is a function whose derivative gives an integrand. This function is called an antiderivative. By this principle and constructs the main integrals.Determine by the form of the integrand which of the table integrals fits in in this case. It is not always possible to determine this immediately. Often, the tabular form becomes noticeable only after several transformations to simplify the integrand.
Variable Replacement Method
If the integrand function is trigonometric function, whose argument contains some polynomial, then try using the variable replacement method. In order to do this, replace the polynomial in the argument of the integrand with some new variable. Based on the relationship between the new and old variables, determine the new limits of integration. By differentiating this expression, find the new differential in . So you will get the new kind of the previous integral, close to or even corresponding to any tabular one.Solving integrals of the second kind
If the integral is an integral of the second kind, a vector form of the integrand, then you will need to use the rules for the transition from these integrals to scalar ones. One such rule is the Ostrogradsky-Gauss relation. This law allows you to go from the rotor flux of some vector function to the triple integral over the divergence of a given vector field.Substitution of integration limits
After finding the antiderivative, it is necessary to substitute the limits of integration. First substitute the value upper limit into an expression for the antiderivative. You will get some number. Next, subtract from the resulting number another number obtained from the lower limit into the antiderivative. If one of the limits of integration is infinity, then when substituting it into antiderivative function it is necessary to go to the limit and find what the expression strives for.If the integral is two-dimensional or three-dimensional, then you will have to represent the limits of integration geometrically to understand how to evaluate the integral. Indeed, in the case of, say, a three-dimensional integral, the limits of integration can be entire planes that limit the volume being integrated.
274. Remarks.
A) If the expression you want to evaluate contains sum or difference numbers, then they must be found without the help of tables ordinary addition or by subtraction. Eg:
log (35 +7.24) 5 = 5 log (35 + 7.24) = 5 log 42.24.
b) Knowing how to logarithm expressions, we can, inversely, by this result using logarithms to find the expression from which this result was obtained; so if
log X= log a+ log b- 3 log With,
then it is easy to understand that
V) Before moving on to considering the structure of logarithmic tables, we will indicate some properties of decimal logarithms, i.e. those in which the number 10 is taken as the base (only such logarithms are used for calculations).
Chapter two.
Properties of decimal logarithms.
275 . A) Since 10 1 = 10, 10 2 = 100, 10 3 = 1000, 10 4 = 10000, etc., then log 10 = 1, log 100 = 2, log 1000 = 3, log 10000 = 4, and etc.
Means, the logarithm of an integer represented by one followed by zeros is an integer positive number, containing as many ones as zeros in the number image.
Thus: log 100,000 = 5, log 1000 000 = 6 , etc.
b) Because
log 0.1 = -l; log 0.01 = - 2; log 0.001 == -3; log 0.0001 = - 4, etc.
Means, logarithm decimal, represented by a unit with preceding zeros, is a negative integer containing as many negative units as there are zeros in the representation of the fraction, including 0 integers.
Thus: log 0.00001= - 5, log 0.000001 = -6, etc.
V) Let's take an integer that is not represented by one and zeros, for example. 35, or a whole number with a fraction, for example. 10.7. The logarithm of such a number cannot be an integer, since raising 10 to a power with an integer exponent (positive or negative), we get 1 with zeros (following 1, or preceding it). Let us now assume that the logarithm of such a number is some fraction a / b . Then we would have equality
But these equalities are impossible, as 10A there are 1s with zeros, whereas degrees 35b And 10,7b by any measure b cannot give 1 followed by zeros. This means that we cannot allow log 35 And log 10.7 were equal to fractions. But from the properties logarithmic function we know () that every positive number has a logarithm; consequently, each of the numbers 35 and 10.7 has its own logarithm, and since it cannot be either an integer number or a fractional number, it is an irrational number and, therefore, cannot be expressed exactly by means of numbers. Irrational logarithms are usually expressed approximately as a decimal fraction with several decimal places. The integer number of this fraction (even if it were “0 integers”) is called characteristic, A fraction- mantissa of the logarithm. If, for example, there is a logarithm 1,5441 , then its characteristic is equal 1 , and the mantissa is 0,5441 .
G) Let's take some integer or mixed number, for example. 623 or 623,57 . The logarithm of such a number consists of a characteristic and a mantissa. It turns out that decimal logarithms have the convenience that we can always find their characteristics by one type of number . To do this, we count how many digits are in a given whole number, or in an integer part mixed number, In our examples of these numbers 3 . Therefore, each of the numbers 623 And 623,57 more than 100 but less than 1000; this means that the logarithm of each of them is greater log 100, i.e. more 2 , but less log 1000, i.e. less 3 (remember that a larger number also has a larger logarithm). Hence, log 623 = 2,..., And log 623.57 = 2,... (dots replace unknown mantissas).
Like this we find:
|
10 < 56,7 < 100 1 < log56,7 < 2 log 56.7 = 1,... |
1000 < 8634 < 10 000 3 < log8634 < 4 log 8634 = 3,... |
Let in general a given integer number, or an integer part of a given mixed number, contain m numbers Since the smallest integer containing m numbers, yes 1 With m - 1 zeros at the end, then (denoting this number N) we can write the inequalities:
and therefore,
m - 1 < log N < m ,
log N = ( m - 1) + positive fraction .
So the characteristic logN = m - 1 .
We see in this way that the characteristic of the logarithm of an integer or mixed number contains as many positive units as there are digits in the integer part of the number minus one.
Having noticed this, we can directly write:
log 7.205 = 0,...; log 83 = 1,...; log 720.4 = 2,... and so on.
d) Let's take several decimal fractions smaller 1 (i.e. having 0 whole): 0,35; 0,07; 0,0056; 0,0008, and so on.
Thus, each of these logarithms is contained between two negative integers that differ by one unit; therefore each of them is equal to the smaller of these negative numbers increased by some positive fraction. For example, log0.0056= -3 + positive fraction. Let's assume that this fraction is 0.7482. Then it means:
log 0.0056 = - 3 + 0.7482 (= - 2.2518).
Amounts such as - 3 + 0,7482 , consisting of a negative integer and a positive decimal fraction, were agreed upon logarithmic calculations abbreviated as follows: 3 ,7482 (This number reads: 3 minus, 7482 ten thousandths.), i.e. they put a minus sign over the characteristic in order to show that it relates only to this characteristic, and not to the mantissa, which remains positive. Thus, from the above table it is clear that
log 0.35 == 1 ,....; log 0.07 = 2,....; log 0.0008 = 4 ,....
Let at all 
. there is a decimal fraction in which before the first significant figure α
costs m
zeros, including 0 integers. Then it is obvious that
- m < log A < - (m- 1).
Since from two integers:- m And - (m- 1) there is less - m , That
log A = - m+ positive fraction,
and therefore the characteristic log A = - m (with a positive mantissa).
Thus, the characteristic of the logarithm of a decimal fraction less than 1 contains as many negative ones as there are zeros in the image of the decimal fraction before the first significant digit, including zero integers; The mantissa of such a logarithm is positive.
e) Let's multiply some number N(integer or fractional - it doesn’t matter) by 10, by 100 by 1000..., in general by 1 with zeros. Let's see how this changes log N. Since the logarithm of the product equal to the sum logarithms of the factors, then
log(N 10) = log N + log 10 = log N + 1;
log(N 100) = log N + log 100 = log N + 2;
log(N 1000) = log N + log 1000 = log N + 3; etc.
When to log N we add some integer, then we can always add this number to the characteristic, and not to the mantissa.
So, if log N = 2.7804, then 2.7804 + 1 = 3.7804; 2.7804 + 2 = 4.7801, etc.;
or if log N = 3.5649, then 3.5649 + 1 = 2.5649; 3.5649 + 2 = 1.5649, etc.
When a number is multiplied by 10, 100, 1000,..., generally by 1 with zeros, the mantissa of the logarithm does not change, and the characteristic increases by as many units as there are zeros in the factor .
Similarly, taking into account that the logarithm of the quotient is equal to the logarithm of the dividend without the logarithm of the divisor, we get:
log N / 10 = log N- log 10 = log N -1;
log N / 100 = log N- log 100 = log N -2;
log N / 1000 = log N- log 1000 = log N -3; and so on.
If we agree, when subtracting an integer from a logarithm, to always subtract this integer from the characteristic and leave the mantissa unchanged, then we can say:
Dividing a number by 1 with zeros does not change the mantissa of the logarithm, but the characteristic decreases by as many units as there are zeros in the divisor.
276. Consequences. From property ( e) the following two corollaries can be deduced:
A) The mantissa of the logarithm of a decimal number does not change when moved to a decimal point , because moving a decimal point is equivalent to multiplying or dividing by 10, 100, 1000, etc. Thus, logarithms of numbers:
0,00423, 0,0423, 4,23, 423
differ only in characteristics, but not in mantissas (provided that all mantissas are positive).
b) Mantissas of numbers having the same significant part, but differing only by zeros at the end, are the same: Thus, the logarithms of numbers: 23, 230, 2300, 23,000 differ only in characteristics.
Comment. From specified properties decimal logarithms, it is clear that we can find the characteristics of the logarithm of an integer and a decimal fraction without the help of tables (this is the great convenience of decimal logarithms); as a result, only one mantissa is placed in logarithmic tables; in addition, since finding the logarithms of fractions is reduced to finding the logarithms of integers (the logarithm of a fraction = the logarithm of the numerator without the logarithm of the denominator), the mantissas of logarithms of only integers are placed in the tables.
Chapter three.
Design and use of four-digit tables.
277. Systems of logarithms. A system of logarithms is a set of logarithms calculated for a number of consecutive integers using the same base. Two systems are used: the system of ordinary or decimal logarithms, in which the number is taken as the base 10 , and a system of so-called natural logarithms, in which an irrational number is taken as the base (for some reasons that are clear in other branches of mathematics) 2,7182818 ... For calculations, decimal logarithms are used, due to the convenience that we indicated when we listed the properties of such logarithms.
Natural logarithms are also called Neperov, named after the inventor of logarithms, a Scottish mathematician Nepera(1550-1617), and decimal logarithms - Briggs named after the professor Brigga(a contemporary and friend of Napier), who first compiled tables of these logarithms.
278. Converting a negative logarithm into one whose mantissa is positive, and the inverse transformation. We have seen that the logarithms of numbers less than 1 are negative. This means that they consist of a negative characteristic and a negative mantissa. Such logarithms can always be transformed so that their mantissa is positive, but the characteristic remains negative. To do this, it is enough to add a positive one to the mantissa, and a negative one to the characteristic (which, of course, does not change the value of the logarithm).
If, for example, we have a logarithm - 2,0873 , then you can write:
- 2,0873 = - 2 - 1 + 1 - 0,0873 = - (2 + 1) + (1 - 0,0873) = - 3 + 0,9127,
or abbreviated:
Conversely, any logarithm with a negative characteristic and a positive mantissa can be turned into a negative one. To do this, it is enough to add a negative one to the positive mantissa, and a positive one to the negative characteristic: so, you can write:
279. Description of four-digit tables. For majority decision practical problems Four-digit tables are quite sufficient, the handling of which is very simple. These tables (with the inscription “logarithms” at the top) are placed at the end of this book, and not most of them (to explain the location) are printed on this page. They contain mantissas
Logarithms.
logarithms of all integers from 1 before 9999 inclusive, calculated to four decimal places, with the last of these places increased by 1 in all those cases where the 5th decimal place would be 5 or more than 5; therefore, 4-digit tables give approximate mantissas up to 1 / 2 ten-thousandth part (with a deficiency or excess).
Since we can directly characterize the logarithm of an integer or a decimal fraction, based on the properties of decimal logarithms, we must take only the mantissas from the tables; At the same time, we must remember that the position of the comma in decimal number, as well as the number of zeros at the end of the number, have no effect on the value of the mantissa. Therefore, when finding the mantissa by given number we discard the comma in this number, as well as the zeros at the end of it, if there are any, and find the mantissa of the integer formed after this. The following cases may arise.
1) An integer consists of 3 digits. For example, let’s say we need to find the mantissa of the logarithm of the number 536. The first two digits of this number, i.e. 53, are found in the tables in the first vertical column on the left (see table). Having found the number 53, we move from it along a horizontal line to the right until this line intersects with a vertical column passing through one of the numbers 0, 1, 2, 3,... 9, placed at the top (and bottom) of the table, which is 3- th digit of a given number, i.e. in our example, the number 6. At the intersection we get the mantissa 7292 (i.e. 0.7292), which belongs to the logarithm of the number 536. Similarly, for the number 508 we find the mantissa 0.7059, for the number 500 we find 0.6990, etc.
2) An integer consists of 2 or 1 digits. Then we mentally assign one or two zeros to this number and find the mantissa for the three-digit number thus formed. For example, we add one zero to the number 51, from which we get 510 and find the mantissa 7070; to the number 5 we assign 2 zeros and find the mantissa 6990, etc.
3) An integer is expressed in 4 digits. For example, you need to find the mantissa of log 5436. Then first we find in the tables, as just indicated, the mantissa for the number represented by the first 3 digits of this number, i.e. for 543 (this mantissa will be 7348); then we move from the found mantissa along the horizontal line to the right (to the right side of the table, located behind the thick vertical line) until it intersects with the vertical column passing through one of the numbers: 1, 2 3,... 9, located at the top (and at the bottom ) of this part of the table, which represents the 4th digit of a given number, i.e., in our example, the number 6. At the intersection we find the correction (number 5), which must be mentally applied to the mantissa of 7348 in order to obtain the mantissa of the number 5436; This way we get the mantissa 0.7353.
4) An integer is expressed with 5 or more digits. Then we discard all digits except the first 4, and take an approximate four-digit number, and increase the last digit of this number by 1 in that number. case when the discarded 5th digit of the number is 5 or more than 5. So, instead of 57842 we take 5784, instead of 30257 we take 3026, instead of 583263 we take 5833, etc. For this rounded four-digit number, we find the mantissa as just explained.
Guided by these guidelines, let us find the logarithms as an example the following numbers:
36,5; 804,7; 0,26; 0,00345; 7,2634; 3456,06.
First of all, without turning to the tables for now, we will put down only the characteristics, leaving room for the mantissas, which we will write out after:
log 36.5 = 1,.... log 0.00345 = 3,....
log 804.7 = 2,.... log 7.2634 = 0,....
log 0.26 = 1,.... log 3456.86 = 3,....
log 36.5 = 1.5623; log 0.00345 = 3.5378;
log 804.7 = 2.9057; log 7.2634 = 0.8611;
log 0.26 = 1.4150; log 3456.86 = 3.5387.
280. Note. In some four-digit tables (for example, in tables V. Lorchenko and N. Ogloblina, S. Glazenap, N. Kamenshchikova) corrections for the 4th digit of this number are not placed. When dealing with such tables, you have to find these corrections using simple calculation, which can be performed on the basis of the following truth: if the numbers exceed 100, and the differences between them are less than 1, then without sensitive error it can be accepted that differences between logarithms are proportional to differences between corresponding numbers . Let, for example, we need to find the mantissa corresponding to the number 5367. This mantissa, of course, is the same as for the number 536.7. We find in the tables for the number 536 the mantissa 7292. Comparing this mantissa with the mantissa 7300 adjacent to the right, corresponding to the number 537, we notice that if the number 536 increases by 1, then its mantissa will increase by 8 ten-thousandths (8 is the so-called table difference between two adjacent mantissas); if the number 536 increases by 0.7, then its mantissa will increase not by 8 ten-thousandths, but by some smaller numberX ten thousandths, which, according to the assumed proportionality, must satisfy the proportions:
X :8 = 0.7:1; where X = 8 07 = 5,6,
which is rounded to 6 ten-thousandths. This means that the mantissa for the number 536.7 (and therefore for the number 5367) will be: 7292 + 6 = 7298.
Note that finding an intermediate number using two adjacent numbers in tables is called interpolation. The interpolation described here is called proportional, since it is based on the assumption that the change in the logarithm is proportional to the change in the number. It is also called linear, since it assumes that graphically the change in a logarithmic function is expressed by a straight line.
281. Error limit of the approximate logarithm. If the number whose logarithm is being sought is an exact number, then the limit of error of its logarithm found in 4-digit tables can, as we said in, be taken 1 / 2 ten-thousandth part. If this number is not exact, then to this error limit we must also add the limit of another error resulting from the inaccuracy of the number itself. It has been proven (we omit this proof) that such a limit can be taken to be the product
a(d +1) ten thousandths.,
in which A is the margin of error for the most imprecise number, assuming that its integer part contains 3 digits, a d tabular difference of mantissas corresponding to two consecutive three-digit numbers between which the given imprecise number lies. Thus, the limit of the final error of the logarithm will then be expressed by the formula:
1 / 2 + a(d +1) ten thousandths
Example. Find log π , taking for π approximate number 3.14, exact to 1 / 2 hundredth.
Moving the comma after the 3rd digit in the number 3.14, counting from the left, we get three digit number 314, exact to 1 / 2 units; This means that the margin of error for an inaccurate number, i.e., what we denoted by the letter A , there is 1 / 2 From the tables we find:
log 3.14 = 0.4969.
Table difference d between the mantissas of the numbers 314 and 315 is equal to 14, so the error of the found logarithm will be less
1 / 2 + 1 / 2 (14 +1) = 8 ten thousandths.
Since we do not know about the logarithm 0.4969 whether it is deficient or excessive, we can only guarantee that the exact logarithm π lies between 0.4969 - 0.0008 and 0.4969 + 0.0008, i.e. 0.4961< log π < 0,4977.
282. Find a number using a given logarithm. To find a number using a given logarithm, the same tables can be used to find the mantissas of given numbers; but it is more convenient to use other tables that contain the so-called antilogarithms, i.e., numbers corresponding to these mantissas. These tables, indicated by the inscription at the top “antilogarithms,” are placed at the end of this book after the tables of logarithms; a small part of them is placed on this page (for explanation).
Suppose you are given a 4-digit mantissa 2863 (we do not pay attention to the characteristic) and you need to find the corresponding integer. Then, having tables of antilogarithms, you need to use them in exactly the same way as was previously explained to find the mantissa for a given number, namely: we find the first 2 digits of the mantissa in the first column on the left. Then we move from these numbers along the horizontal line to the right until it intersects with the vertical column coming from the 3rd digit of the mantissa, which must be looked for in the top line (or bottom). At the intersection we find the four-digit number 1932, corresponding to the mantissa 286. Then from this number we move further along the horizontal line to the right until the intersection with the vertical column coming from the 4th digit of the mantissa, which must be found at the top (or bottom) among the numbers 1, 2 placed there , 3,... 9. At the intersection we find correction 1, which must be applied (in the mind) to the number 1032 found earlier in order to obtain the number corresponding to the mantissa 2863.
Thus, the number will be 1933. After this, paying attention to the characteristic, you need to put occupied in the proper place in the number 1933. For example:
If log x = 3.2863, then X = 1933,
„ log x = 1,2863, „ X = 19,33,
, log x = 0,2&63, „ X = 1,933,
„ log x = 2 ,2863, „ X = 0,01933
Here are more examples:
log x = 0,2287, X = 1,693,
log x = 1 ,7635, X = 0,5801,
log x = 3,5029, X = 3184,
log x = 2 ,0436, X = 0,01106.
If the mantissa contains 5 or more digits, then we take only the first 4 digits, discarding the rest (and increasing the 4th digit by 1 if the 5th digit has five or more). For example, instead of the mantissa 35478 we take 3548, instead of 47562 we take 4756.
283. Note. The correction for the 4th and subsequent digits of the mantissa can also be found through interpolation. So, if the mantissa is 84357, then, having found the number 6966, corresponding to the mantissa 843, we can further reason as follows: if the mantissa increases by 1 (thousandth), i.e., it makes 844, then the number, as can be seen from the tables, will increase by 16 units; if the mantissa increases not by 1 (thousandth), but by 0.57 (thousandth), then the number will increase by X units, and X must satisfy the proportions:
X : 16 = 0.57: 1, from where x = 16 0,57 = 9,12.
This means that the required number will be 6966+ 9.12 = 6975.12 or (limited to only four digits) 6975.
284. Error limit of the found number. It has been proven that in the case when in the found number the comma is after the 3rd digit from the left, i.e. when the characteristic of the logarithm is 2, the sum can be taken as the error limit
Where A is the error limit of the logarithm (expressed in ten thousandths) by which the number was found, and d - the difference between the mantissas of two three-digit consecutive numbers between which the found number lies (with a comma after the 3rd digit from the left). When the characteristic is not 2, but some other, then in the found number the comma will have to be moved to the left or to the right, i.e., divide or multiply the number by some power of 10. In this case, the error of the result will also be divided or multiplied by the same power of 10.
Let, for example, we are looking for a number using the logarithm 1,5950 , which is known to be accurate to 3 ten-thousandths; that means then A = 3 . The number corresponding to this logarithm, found from the table of antilogarithms, is 39,36 . Moving the comma after the 3rd digit from the left, we have the number 393,6 , consisting between 393 And 394 . From the tables of logarithms we see that the difference between the mantissas corresponding to these two numbers is 11 ten thousandths; Means d = 11 . The error of the number 393.6 will be less
This means that the error in the number 39,36 there will be less 0,05 .
285. Operations on logarithms with negative characteristics. Adding and subtracting logarithms does not present any difficulties, as can be seen from the following examples:
There is also no difficulty in multiplying the logarithm by a positive number, for example:
IN last example separately multiply the positive mantissa by 34, then negative characteristic at 34.
If the logarithm of a negative characteristic and a positive mantissa is multiplied by a negative number, then proceed in two ways: either the given logarithm is first turned negative, or the mantissa and characteristic are multiplied separately and the results are combined together, for example:
3 ,5632 (- 4) = - 2,4368 (- 4) = 9,7472;
3 ,5632 (- 4) = + 12 - 2,2528 = 9,7472.
When dividing, two cases may arise: 1) the negative characteristic is divided and 2) is not divisible by a divisor. In the first case, the characteristic and mantissa are separated separately:
10 ,3784: 5 = 2 ,0757.
In the second case, so many negative units are added to the characteristic so that the resulting number is divided by the divisor; the same number of positive units is added to the mantissa:
3 ,7608: 8 = (- 8 + 5,7608) : 8 = 1 ,7201.
This transformation must be done in the mind, so the action goes like this:
286. Replacing subtracted logarithms with terms. When calculating some complex expression using logarithms you have to add some logarithms, subtract others; in this case, in the usual way of performing actions, they separately find the sum of the added logarithms, then the sum of the subtracted ones, and subtract the second from the first sum. For example, if we have:
log X = 2,7305 - 2 ,0740 + 3 ,5464 - 8,3589 ,
then the usual execution of actions will look like this:
However, it is possible to replace subtraction with addition. So:
Now you can arrange the calculation like this:
287. Examples of calculations.
Example 1. Evaluate expression:
If A = 0.8216, B = 0.04826, C = 0.005127 And D = 7.246.
Let's take logarithm this expression:
log X= 1/3 log A + 4 log B - 3 log C - 1/3 log D
Now, to avoid unnecessary loss of time and to reduce the possibility of errors, first of all we will arrange all the calculations without executing them for now and, therefore, without referring to the tables:
After this, we take the tables and put logarithms on the remaining free places:
Error limit. First, let's find the limit of error of the number x 1 = 194,5 , equal to:
So, first of all you need to find A , i.e., the error limit of the approximate logarithm, expressed in ten thousandths. Let us assume that these numbers A, B, C And D all are accurate. Then the errors in individual logarithms will be as follows (in ten thousandths):
V logA.......... 1 / 2
V 1/3 log A......... 1 / 6 + 1 / 2 = 2 / 3
( 1 / 2 added because when dividing by 3 logarithms of 1.9146, we rounded the quotient by discarding its 5th digit, and, therefore, made an even smaller error 1 / 2 ten-thousandth).
Now we find the error limit of the logarithm:
A = 2 / 3 + 2 + 3 / 2 + 1 / 6 = 4 1 / 3 (ten thousandths).
Let us further define d . Because x 1 = 194,5 , then 2 integers consecutive numbers, between which lies x 1 will 194 And 195 . Table difference d between the mantissas corresponding to these numbers is equal to 22 . This means that the limit of error of the number is x 1 There is:
Because x = x 1 : 10, then the error limit in the number x equals 0,3:10 = 0,03 . Thus, the number we found 19,45 differs from the exact number by less than 0,03 . Since we do not know whether our approximation was found with a deficiency or with an excess, we can only guarantee that
19,45 + 0,03 > X > 19,45 - 0,03 , i.e.
19,48 > X > 19,42 ,
and therefore, if we accept X =19,4 , then we will have an approximation with a disadvantage with an accuracy of up to 0.1.
Example 2. Calculate:
X = (- 2,31) 3 5 √72 = - (2,31) 3 5 √72 .
Because negative numbers do not have logarithms, then we first find:
X" = (2,31) 3 5 √72
by decomposition:
log X"= 3 log 2.31 + 1 / 5 log72.
After calculation it turns out:
X" = 28,99 ;
hence,
x = - 28,99 .
Example 3. Calculate:
Continuous logarithmization cannot be used here, since the sign of the root is c u m m a. In such cases, calculate the formula by parts.
First we find N = 5 √8 , Then N 1 = 4 √3 ; then by simple addition we determine N+ N 1 , and finally we calculate 3 √N+ N 1 ; it turns out:
N=1.514, N 1 = 1,316 ; N+ N 1 = 2,830 .
log x= log 3 √ 2,830 = 1 / 3 log 2.830 = 0,1506 ;
x = 1,415 .
Chapter Four.
Exponential and logarithmic equations.
288. Exponential equations are those in which the unknown is included in the exponent, and logarithmic- those in which the unknown enters under the sign log. Such equations can be solvable only in special cases, and one has to rely on the properties of logarithms and on the principle that if the numbers are equal, then their logarithms are equal, and, conversely, if the logarithms are equal, then the corresponding numbers are equal.
Example 1. Solve the equation: 2 x = 1024 .
Let's logarithm both sides of the equation:
Example 2. Solve the equation: a 2x - a x = 1 . Putting a x = at , we get a quadratic equation:
y 2 - at - 1 = 0 ,
Because 1-√5 < 0 , then the last equation is impossible (function a x there is always a positive number), and the first gives:
Example 3. Solve the equation:
log( a + x) + log ( b + x) = log ( c + x) .
The equation can be written like this:
log [( a + x) (b + x)] = log ( c + x) .
From the equality of logarithms we conclude that the numbers are equal:
(a + x) (b + x) = c + x .
This is a quadratic equation, the solution of which is not difficult.
Chapter five.
Compound interest, term payments and term payments.
289. Basic problem on compound interest. How much will the capital turn into? A rubles, given in growth at R compound interest, after the lapse of t years ( t - integer)?
They say that capital is paid at compound interest if the so-called “interest on interest” is taken into account, that is, if the interest money due on the capital is added to the capital at the end of each year in order to increase it with interest in subsequent years.
Every ruble of capital given away R %, will bring profit within one year p / 100 ruble, and, therefore, every ruble of capital in 1 year will turn into 1 + p / 100 ruble (for example, if capital is given at 5 %, then every ruble of it in a year will turn into 1 + 5 / 100 , i.e. in 1,05 ruble).
For brevity, denoting the fraction p / 100 with one letter, for example, r , we can say that every ruble of capital in a year will turn into 1 + r rubles; hence, A rubles will be returned in 1 year to A (1 + r ) rub. After another year, i.e. 2 years from the start of growth, every ruble of these A (1 + r ) rub. will contact again 1 + r rub.; This means that all capital will turn into A (1 + r ) 2 rub. In the same way we find that after three years the capital will be A (1 + r ) 3 , in four years it will be A (1 + r ) 4 ,... generally through t years if t is an integer, it will turn to A (1 + r ) t rub. Thus, denoting by A final capital, we will have the following formula compound interest:
A = A (1 + r ) t Where r = p / 100 .
Example. Let a =2,300 rub., p = 4, t=20 years; then the formula gives:
r = 4 / 100 = 0,04 ; A = 2,300 (1.04) 20.
To calculate A, we use logarithms:
log a = log 2 300 + 20 log 1.04 = 3.3617 + 20 0.0170 = 3.3617+0.3400 = 3.7017.
A = 5031 ruble.
Comment. In this example we had to log 1.04 multiply by 20 . Since the number 0,0170 there is an approximate value log 1.04 up to 1 / 2 ten-thousandth part, then the product of this number by 20 it will definitely only be until 1 / 2 20, i.e. up to 10 ten-thousandths = 1 thousandth. Therefore in total 3,7017 We cannot vouch not only for the number of ten thousandths, but also for the number of thousandths. So that in such cases it is possible to obtain greater accuracy, better for number 1 + r take logarithms not 4-digit, but with a large number numbers, eg. 7-digit. For this purpose, we present here a small table in which 7-digit logarithms are written out for the most common values R .
290. The main task is for urgent payments. Someone took A rubles per R % with the condition to repay the debt, together with the interest due on it, in t years, paying the same amount at the end of each year. What should this amount be?
Sum x , paid annually under such conditions, is called urgent payment. Let us again denote by the letter r annual interest money from 1 rub., i.e. the number p / 100 . Then by the end of the first year the debt A increases to A (1 + r ), basic payment X it will cost rubles A (1 + r )-X .
By the end of the second year, every ruble of this amount will again turn into 1 + r rubles, and therefore the debt will be [ A (1 + r )-X ](1 + r ) = A (1 + r ) 2 - x (1 + r ), and for payment x rubles will be: A (1 + r ) 2 - x (1 + r ) - X . In the same way, we will make sure that by the end of the 3rd year the debt will be
A (1 + r ) 3 - x (1 + r ) 2 - x (1 + r ) - x ,
and in general and the end t year it will be:
A (1 + r ) t - x (1 + r ) t -1 - x (1 + r ) t -2 ... - x (1 + r ) - x , or
A (1 + r ) t - x [ 1 + (1 + r ) + (1 + r ) 2 + ...+ (1 + r ) t -2 + (1 + r ) t -1 ]
The polynomial inside the parentheses represents the sum of the terms geometric progression; which has the first member 1 , last ( 1 + r ) t -1, and the denominator ( 1 + r ). Using the formula for the sum of terms of a geometric progression (Section 10, Chapter 3, § 249), we find:
and the amount of debt after t -th payment will be:
According to the conditions of the problem, the debt is at the end t -th year must be equal to 0 ; That's why:
where
When calculating this urgent payment formulas using logarithms we must first find the auxiliary number N = (1 + r ) t by logarithm: log N= t log(1+ r) ; having found N, subtract 1 from it, then we get the denominator of the formula for X, after which we find by secondary logarithm:
log X= log a+ log N + log r - log (N - 1).
291. The main task for term contributions. Someone deposits the same amount into the bank at the beginning of each year. A rub. Determine what capital will be formed from these contributions after t years if the bank pays R compound interest.
Designated by r annual interest money from 1 ruble, i.e. p / 100 , we reason like this: by the end of the first year the capital will be A (1 + r );
at the beginning of the 2nd year will be added to this amount A rubles; this means that at this time capital will be A (1 + r ) + a . By the end of the 2nd year he will be A (1 + r ) 2 + a (1 + r );
at the beginning of the 3rd year it is entered again A rubles; this means that at this time there will be capital A (1 + r ) 2 + a (1 + r ) + A ; by the end of the 3rd he will be A (1 + r ) 3 + a (1 + r ) 2 + a (1 + r ) Continuing these arguments further, we find that by the end t year the required capital A will:
This is the formula for term contributions made at the beginning of each year.
The same formula can be obtained by the following reasoning: down payment to A rubles while in the bank t years, will turn, according to the compound interest formula, into A (1 + r ) t rub. The second installment, being in the bank for one year less, i.e. t - 1 years old, contact A (1 + r ) t- 1 rub. Likewise, the third installment will give A (1 + r ) t-2 etc., and finally the last installment, having been in the bank for only 1 year, will go to A (1 + r ) rub. This means the final capital A rub. will:
A= A (1 + r ) t + A (1 + r ) t- 1 + A (1 + r ) t-2 + . . . + A (1 + r ),
which, after simplification, gives the formula found above.
When calculating using logarithms of this formula, you must proceed in the same way as when calculating the formula for urgent payments, i.e., first find the number N = ( 1 + r ) t by its logarithm: log N= t log(1 + r ), then the number N- 1 and then take a logarithm of the formula:
log A = log a+log(1+ r) + log (N - 1) - 1ogr
Comment. If an urgent contribution to A rub. was made not at the beginning, but at the end of each year (as, for example, an urgent payment is made X to pay off the debt), then, reasoning similarly to the previous one, we find that by the end t year the required capital A" rub. will be (including the last installment A rub., not bearing interest):
A"= A (1 + r ) t- 1 + A (1 + r ) t-2 + . . . + A (1 + r ) + A
which is equal to:
i.e. A" ends up in ( 1 + r ) times less A, which was to be expected, since every ruble of capital A" lies in the bank for a year less than the corresponding ruble of capital A.