One of the elements of primitive level algebra is the logarithm. The name comes from the Greek language from the word “number” or “power” and means the power to which the number in the base must be raised to find the final number.
Types of logarithms
- log a b – logarithm of the number b to base a (a > 0, a ≠ 1, b > 0);
- log b – decimal logarithm (logarithm to base 10, a = 10);
- ln b – natural logarithm (logarithm to base e, a = e).
How to solve logarithms?
The logarithm of b to base a is an exponent that requires b to be raised to base a. The result obtained is pronounced like this: “logarithm of b to base a.” The solution to logarithmic problems is that you need to determine the given power in numbers from the specified numbers. There are some basic rules to determine or solve the logarithm, as well as convert the notation itself. Using them, logarithmic equations are solved, derivatives are found, integrals are solved, and many other operations are carried out. Basically, the solution to the logarithm itself is its simplified notation. Below are the basic formulas and properties:
For any a ; a > 0; a ≠ 1 and for any x ; y > 0.
- a log a b = b – basic logarithmic identity
- log a 1 = 0
- loga a = 1
- log a (x y) = log a x + log a y
- log a x/ y = log a x – log a y
- log a 1/x = -log a x
- log a x p = p log a x
- log a k x = 1/k log a x , for k ≠ 0
- log a x = log a c x c
- log a x = log b x/ log b a – formula for moving to a new base
- log a x = 1/log x a
How to solve logarithms - step-by-step instructions for solving
- First, write down the required equation.
Please note: if the base logarithm is 10, then the entry is shortened, resulting in a decimal logarithm. If there is a natural number e, then we write it down, reducing it to a natural logarithm. This means that the result of all logarithms is the power to which the base number is raised to obtain the number b.
Directly, the solution lies in calculating this degree. Before solving an expression with a logarithm, it must be simplified according to the rule, that is, using formulas. You can find the main identities by going back a little in the article.
When adding and subtracting logarithms with two different numbers but with the same bases, replace with one logarithm with the product or division of the numbers b and c, respectively. In this case, you can apply the formula for moving to another base (see above).
If you use expressions to simplify a logarithm, there are some limitations to consider. And that is: the base of the logarithm a is only a positive number, but not equal to one. The number b, like a, must be greater than zero.
There are cases where, by simplifying an expression, you will not be able to calculate the logarithm numerically. It happens that such an expression does not make sense, because many powers are irrational numbers. Under this condition, leave the power of the number as a logarithm.
Follows from its definition. And so the logarithm of the number b based on A is defined as the exponent to which a number must be raised a to get the number b(logarithm exists only for positive numbers).
From this formulation it follows that the calculation x=log a b, is equivalent to solving the equation a x =b. For example, log 2 8 = 3 because 8 = 2 3 . The formulation of the logarithm makes it possible to justify that if b=a c, then the logarithm of the number b based on a equals With. It is also clear that the topic of logarithms is closely related to the topic of powers of a number.
With logarithms, as with any numbers, you can do operations of addition, subtraction and transform in every possible way. But due to the fact that logarithms are not entirely ordinary numbers, their own special rules apply here, which are called main properties.
Adding and subtracting logarithms.
Let's take two logarithms with the same bases: log a x And log a y. Then it is possible to perform addition and subtraction operations:
log a x+ log a y= log a (x·y);
log a x - log a y = log a (x:y).
log a(x 1 . x 2 . x 3 ... x k) = log a x 1 + log a x 2 + log a x 3 + ... + log a x k.
From logarithm quotient theorem One more property of the logarithm can be obtained. It is common knowledge that log a 1= 0, therefore
log a 1 /b=log a 1 - log a b= -log a b.
This means there is an equality:
log a 1 / b = - log a b.
Logarithms of two reciprocal numbers for the same reason will differ from each other solely by sign. So:
Log 3 9= - log 3 1 / 9 ; log 5 1 / 125 = -log 5 125.
Problem B7 gives some expression that needs to be simplified. The result should be a regular number that can be written down on your answer sheet. All expressions are conventionally divided into three types:
- Logarithmic,
- Indicative,
- Combined.
Exponential and logarithmic expressions in their pure form are practically never found. However, knowing how they are calculated is absolutely necessary.
In general, problem B7 is solved quite simply and is quite within the capabilities of the average graduate. The lack of clear algorithms is compensated for by its standardization and monotony. You can learn to solve such problems simply through a lot of training.
Logarithmic Expressions
The vast majority of B7 problems involve logarithms in one form or another. This topic is traditionally considered difficult, since its study usually occurs in the 11th grade - the era of mass preparation for final exams. As a result, many graduates have a very vague understanding of logarithms.
But in this task no one requires deep theoretical knowledge. We will encounter only the simplest expressions that require simple reasoning and can easily be mastered independently. Below are the basic formulas you need to know to cope with logarithms:
In addition, you must be able to replace roots and fractions with powers with a rational exponent, otherwise in some expressions there will simply be nothing to take out from under the logarithm sign. Replacement formulas:
Task. Find the meaning of expressions:
log 6 270 − log 6 7.5
log 5 775 − log 5 6.2
The first two expressions are converted as the difference of logarithms:
log 6 270 − log 6 7.5 = log 6 (270: 7.5) = log 6 36 = 2;
log 5 775 − log 5 6.2 = log 5 (775: 6.2) = log 5 125 = 3.
To calculate the third expression, you will have to isolate powers - both in the base and in the argument. First, let's find the internal logarithm:
Then - external:
Constructions of the form log a log b x seem complex and misunderstood to many. Meanwhile, this is just a logarithm of the logarithm, i.e. log a (log b x ). First, the internal logarithm is calculated (put log b x = c), and then the external one: log a c.
Demonstrative Expressions
We will call an exponential expression any construction of the form a k, where the numbers a and k are arbitrary constants, and a > 0. Methods for working with such expressions are quite simple and are discussed in 8th grade algebra lessons.
Below are the basic formulas that you definitely need to know. The application of these formulas in practice, as a rule, does not cause problems.
- a n · a m = a n + m ;
- a n / a m = a n − m ;
- (a n ) m = a n · m ;
- (a · b ) n = a n · b n ;
- (a : b ) n = a n : b n .
If you come across a complex expression with powers, and it is not clear how to approach it, use a universal technique - decomposition into simple factors. As a result, large numbers in the bases of powers are replaced by simple and understandable elements. Then all that remains is to apply the above formulas - and the problem will be solved.
Task. Find the values of the expressions: 7 9 · 3 11: 21 8, 24 7: 3 6: 16 5, 30 6: 6 5: 25 2.
Solution. Let's decompose all the bases of powers into simple factors:
7 9 3 11: 21 8 = 7 9 3 11: (7 3) 8 = 7 9 3 11: (7 8 3 8) = 7 9 3 11: 7 8: 3 8 = 7 3 3 = 189.
24 7: 3 6: 16 5 = (3 2 3) 7: 3 6: (2 4) 5 = 3 7 2 21: 3 6: 2 20 = 3 2 = 6.
30 6: 6 5: 25 2 = (5 3 2) 6: (3 2) 5: (5 2) 2 = 5 6 3 6 2 6: 3 5: 2 5: 5 4 = 5 2 3 2 = 150.
Combined tasks
If you know the formulas, then all exponential and logarithmic expressions can be solved literally in one line. However, in Problem B7 powers and logarithms can be combined to form quite strong combinations.
The basic properties of the natural logarithm, graph, domain of definition, set of values, basic formulas, derivative, integral, power series expansion and representation of the function ln x using complex numbers are given.
Definition
Natural logarithm is the function y = ln x, the inverse of the exponential, x = e y, and is the logarithm to the base of 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 the exponential graph by mirror reflection relative to the straight line y = x.
The natural logarithm is defined for positive values of the 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 with a positive exponent 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 substitution formula:
Proofs of these formulas are presented in the section "Logarithm".
Inverse function
The inverse of the natural logarithm is the 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 by 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.
Tasks whose solution is converting logarithmic expressions, are quite common on the Unified State Examination.
In order to successfully cope with them with minimal time, in addition to the basic logarithmic identities, you need to know and correctly use some more formulas.
This is: a log a b = b, where a, b > 0, a ≠ 1 (It follows directly from the definition of the logarithm).
log a b = log c b / log c a or log a b = 1/log b a
where a, b, c > 0; a, c ≠ 1.
log a m b n = (m/n) log |a| |b|
where a, b > 0, a ≠ 1, m, n Є R, n ≠ 0.
a log c b = b log c a
where a, b, c > 0 and a, b, c ≠ 1
To show the validity of the fourth equality, let’s take the logarithm of the left and right sides to base a. We get log a (a log with b) = log a (b log with a) or log with b = log with a · log a b; log c b = log c a · (log c b / log c a); log with b = log with b.
We have proven the equality of logarithms, which means that the expressions under the logarithms are also equal. Formula 4 has been proven.
Example 1.
Calculate 81 log 27 5 log 5 4 .
Solution.
81 = 3 4 , 27 = 3 3 .
log 27 5 = 1/3 log 3 5, log 5 4 = log 3 4 / log 3 5. Therefore,
log 27 5 log 5 4 = 1/3 log 3 5 (log 3 4 / log 3 5) = 1/3 log 3 4.
Then 81 log 27 5 log 5 4 = (3 4) 1/3 log 3 4 = (3 log 3 4) 4/3 = (4) 4/3 = 4 3 √4.
You can complete the following task yourself.
Calculate (8 log 2 3 + 3 1/ log 2 3) - log 0.2 5.
As a hint, 0.2 = 1/5 = 5 -1 ; log 0.2 5 = -1.
Answer: 5.
Example 2.
Calculate (√11) log √3 9- log 121 81 .
Solution.
Let's change the expressions: 9 = 3 2, √3 = 3 1/2, log √3 9 = 4,
121 = 11 2, 81 = 3 4, log 121 81 = 2 log 11 3 (formula 3 was used).
Then (√11) log √3 9- log 121 81 = (11 1/2) 4-2 log 11 3 = (11) 2- log 11 3 = 11 2 / (11) log 11 3 = 11 2 / ( 11 log 11 3) = 121/3.
Example 3.
Calculate log 2 24 / log 96 2 - log 2 192 / log 12 2.
Solution.
We replace the logarithms contained in the example with logarithms with base 2.
log 96 2 = 1/log 2 96 = 1/log 2 (2 5 3) = 1/(log 2 2 5 + log 2 3) = 1/(5 + log 2 3);
log 2 192 = log 2 (2 6 3) = (log 2 2 6 + log 2 3) = (6 + log 2 3);
log 2 24 = log 2 (2 3 3) = (log 2 2 3 + log 2 3) = (3 + log 2 3);
log 12 2 = 1/log 2 12 = 1/log 2 (2 2 3) = 1/(log 2 2 2 + log 2 3) = 1/(2 + log 2 3).
Then log 2 24 / log 96 2 – log 2 192 / log 12 2 = (3 + log 2 3) / (1/(5 + log 2 3)) – ((6 + log 2 3) / (1/( 2 + log 2 3)) =
= (3 + log 2 3) · (5 + log 2 3) – (6 + log 2 3)(2 + log 2 3).
After opening the parentheses and bringing similar terms, we get the number 3. (When simplifying the expression, we can denote log 2 3 by n and simplify the expression
(3 + n) · (5 + n) – (6 + n)(2 + n)).
Answer: 3.
You can complete the following task yourself:
Calculate (log 3 4 + log 4 3 + 2) log 3 16 log 2 144 3.
Here it is necessary to make the transition to base 3 logarithms and factorization of large numbers into prime factors.
Answer:1/2
Example 4.
Given three numbers A = 1/(log 3 0.5), B = 1/(log 0.5 3), C = log 0.5 12 – log 0.5 3. Arrange them in ascending order.
Solution.
Let's transform the numbers A = 1/(log 3 0.5) = log 0.5 3; C = log 0.5 12 – log 0.5 3 = log 0.5 12/3 = log 0.5 4 = -2.
Let's compare them
log 0.5 3 > log 0.5 4 = -2 and log 0.5 3< -1 = log 0,5 2, так как функция у = log 0,5 х – убывающая.
Or 2< log 0,5 3 < -1. Тогда -1 < 1/(log 0,5 3) < -1/2.
Answer. Therefore, the order of placing the numbers is: C; A; IN.
Example 5.
How many integers are in the interval (log 3 1 / 16 ; log 2 6 48).
Solution.
Let us determine between which powers of the number 3 the number 1/16 is located. We get 1/27< 1 / 16 < 1 / 9 .
Since the function y = log 3 x is increasing, then log 3 (1 / 27)< log 3 (1 / 16) < log 3 (1 / 9); -3 < log 3 (1 / 16) < -2.
log 6 48 = log 6 (36 4 / 3) = log 6 36 + log 6 (4 / 3) = 2 + log 6 (4 / 3). Let's compare log 6 (4 / 3) and 1 / 5. And for this we compare the numbers 4/3 and 6 1/5. Let's raise both numbers to the 5th power. We get (4 / 3) 5 = 1024 / 243 = 4 52 / 243< 6. Следовательно,
log 6 (4 / 3)< 1 / 5 . 2 < log 6 48 < 2 1 / 5 . Числа, входящие в двойное неравенство, положительные. Их можно возводить в квадрат. Знаки неравенства при этом не изменятся. Тогда 4 < log 6 2 48 < 4 21 / 25.
Therefore, the interval (log 3 1 / 16 ; log 6 48) includes the interval [-2; 4] and the integers -2 are placed on it; -1; 0; 1; 2; 3; 4.
Answer: 7 integers.
Example 6.
Calculate 3 lglg 2/ lg 3 - lg20.
Solution.
3 lg lg 2/ lg 3 = (3 1/ lg3) lg lg 2 = (3 lо g 3 10) lg lg 2 = 10 lg lg 2 = lg2.
Then 3 lglg2/lg3 - lg 20 = lg 2 – lg 20 = lg 0.1 = -1.
Answer: -1.
Example 7.
It is known that log 2 (√3 + 1) + log 2 (√6 – 2) = A. Find log 2 (√3 –1) + log 2 (√6 + 2).
Solution.
Numbers (√3 + 1) and (√3 – 1); (√6 – 2) and (√6 + 2) are conjugate.
Let us carry out the following transformation of expressions
√3 – 1 = (√3 – 1) · (√3 + 1)) / (√3 + 1) = 2/(√3 + 1);
√6 + 2 = (√6 + 2) · (√6 – 2)) / (√6 – 2) = 2/(√6 – 2).
Then log 2 (√3 – 1) + log 2 (√6 + 2) = log 2 (2/(√3 + 1)) + log 2 (2/(√6 – 2)) =
Log 2 2 – log 2 (√3 + 1) + log 2 2 – log 2 (√6 – 2) = 1 – log 2 (√3 + 1) + 1 – log 2 (√6 – 2) =
2 – log 2 (√3 + 1) – log 2 (√6 – 2) = 2 – A.
Answer: 2 – A.
Example 8.
Simplify and find the approximate value of the expression (log 3 2 log 4 3 log 5 4 log 6 5 ... log 10 9.
Solution.
Let us reduce all logarithms to a common base 10.
(log 3 2 log 4 3 log 5 4 log 6 5 ... log 10 9 = (lg 2 / lg 3) (lg 3 / lg 4) (lg 4 / lg 5) (lg 5 / lg 6) · … · (lg 8 / lg 9) · lg 9 = lg 2 ≈ 0.3010 (The approximate value of lg 2 can be found using a table, slide rule or calculator).
Answer: 0.3010.
Example 9.
Calculate log a 2 b 3 √(a 11 b -3) if log √ a b 3 = 1. (In this example, a 2 b 3 is the base of the logarithm).
Solution.
If log √ a b 3 = 1, then 3/(0.5 log a b = 1. And log a b = 1/6.
Then log a 2 b 3√(a 11 b -3) = 1/2 log a 2 b 3 (a 11 b -3) = log a (a 11 b -3) / (2log a (a 2 b 3) ) = (log a a 11 + log a b -3) / (2(log a a 2 + log a b 3)) = (11 – 3log a b) / (2(2 + 3log a b)) Considering that that log a b = 1/6 we get (11 – 3 1 / 6) / (2(2 + 3 1 / 6)) = 10.5/5 = 2.1.
Answer: 2.1.
You can complete the following task yourself:
Calculate log √3 6 √2.1 if log 0.7 27 = a.
Answer: (3 + a) / (3a).
Example 10.
Calculate 6.5 4/ log 3 169 · 3 1/ log 4 13 + log125.
Solution.
6.5 4/ log 3 169 · 3 1/ log 4 13 + log 125 = (13/2) 4/2 log 3 13 · 3 2/ log 2 13 + 2log 5 5 3 = (13/2) 2 log 13 3 3 2 log 13 2 + 6 = (13 log 13 3 / 2 log 13 3) 2 (3 log 13 2) 2 + 6 = (3/2 log 13 3) 2 (3 log 13 2) 2 + 6 = (3 2 /(2 log 13 3) 2) · (2 log 13 3) 2 + 6.
(2 log 13 3 = 3 log 13 2 (formula 4))
We get 9 + 6 = 15.
Answer: 15.
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