The logarithm of a positive number b to base a (a>0, a is not equal to 1) is a number c such that a c = b: log a b = c ⇔ a c = b (a > 0, a ≠ 1, b > 0)
Note that the logarithm of a non-positive number is undefined. In addition, the base of the logarithm must be positive number, not equal to 1. For example, if we square -2, we get the number 4, but this does not mean that the logarithm to the base -2 of 4 is equal to 2.
Basic logarithmic identity
a log a b = b (a > 0, a ≠ 1) (2)It is important that the scope of definition of the right and left sides of this formula is different. Left side is defined only for b>0, a>0 and a ≠ 1. The right-hand side is defined for any b, and does not depend on a at all. Thus, the application of the basic logarithmic “identity” when solving equations and inequalities can lead to a change in the OD.
Two obvious consequences of the definition of logarithm
log a a = 1 (a > 0, a ≠ 1) (3)log a 1 = 0 (a > 0, a ≠ 1) (4)
Indeed, when raising the number a to the first power, we get the same number, and when raising it to the first power zero degree- one.
Logarithm of the product and logarithm of the quotient
log a (b c) = log a b + log a c (a > 0, a ≠ 1, b > 0, c > 0) (5)Log a b c = log a b − log a c (a > 0, a ≠ 1, b > 0, c > 0) (6)
I would like to warn schoolchildren against thoughtlessly applying these formulas when solving logarithmic equations and inequalities. When using them “from left to right,” the ODZ narrows, and when moving from the sum or difference of logarithms to the logarithm of the product or quotient, the ODZ expands.
Indeed, the expression log a (f (x) g (x)) is defined in two cases: when both functions are strictly positive or when f(x) and g(x) are both less than zero.
Transforming this expression into the sum log a f (x) + log a g (x) , we are forced to limit ourselves only to the case when f(x)>0 and g(x)>0. There is a narrowing of the area acceptable values, and this is categorically unacceptable, because it can lead to loss of solutions. Similar problem exists for formula (6).
The degree can be taken out of the sign of the logarithm
log a b p = p log a b (a > 0, a ≠ 1, b > 0) (7)And again I would like to call for accuracy. Consider the following example:
Log a (f (x) 2 = 2 log a f (x)
The left side of the equality is obviously defined for all values of f(x) except zero. The right side is only for f(x)>0! By taking the degree out of the logarithm, we again narrow the ODZ. The reverse procedure leads to an expansion of the range of acceptable values. All these remarks apply not only to power 2, but also to any even power.
Formula for moving to a new foundation
log a b = log c b log c a (a > 0, a ≠ 1, b > 0, c > 0, c ≠ 1) (8)That rare case, when the ODZ does not change during the transformation. If you have chosen base c wisely (positive and not equal to 1), the formula for moving to a new base is completely safe.
If we choose the number b as the new base c, we get an important special case formulas (8):
Log a b = 1 log b a (a > 0, a ≠ 1, b > 0, b ≠ 1) (9)
Some simple examples with logarithms
Example 1. Calculate: log2 + log50.
Solution. log2 + log50 = log100 = 2. We used the sum of logarithms formula (5) and the definition of the decimal logarithm.
Example 2. Calculate: lg125/lg5.
Solution. log125/log5 = log 5 125 = 3. We used the formula for moving to a new base (8).
Table of formulas related to logarithms
| a log a b = b (a > 0, a ≠ 1) |
| log a a = 1 (a > 0, a ≠ 1) |
| log a 1 = 0 (a > 0, a ≠ 1) |
| log a (b c) = log a b + log a c (a > 0, a ≠ 1, b > 0, c > 0) |
| log a b c = log a b − log a c (a > 0, a ≠ 1, b > 0, c > 0) |
| log a b p = p log a b (a > 0, a ≠ 1, b > 0) |
| log a b = log c b log c a (a > 0, a ≠ 1, b > 0, c > 0, c ≠ 1) |
| log a b = 1 log b a (a > 0, a ≠ 1, b > 0, b ≠ 1) |
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.
main properties.
- logax + logay = loga(x y);
- logax − logay = loga (x: y).
identical grounds
Log6 4 + log6 9.
Now let's complicate the task a little.
Examples of solving logarithms
What if the base or argument of a logarithm is a power? Then the exponent of this degree can be taken out of the sign of the logarithm according to the following rules:
Of course, all these rules make sense if the ODZ of the logarithm is observed: a > 0, a ≠ 1, x >
Task. Find the meaning of the expression:
Transition to a new foundation
Let it be given logarithm log ax. Then for any number c such that c > 0 and c ≠ 1, the equality is true:
Task. Find the meaning of the expression:
See also:
Basic properties of the logarithm
1.
2.
3.
4.
5. 
6.
7.
8.
9.
10.
11. 
12.
13.
14. 
15.
The exponent is 2.718281828…. To remember the exponent, you can study the rule: the exponent is equal to 2.7 and twice the year of birth of Leo Nikolaevich Tolstoy.
Basic properties of logarithms
Knowing this rule, you will know and exact value exhibitors, and the date of birth of Leo Tolstoy.
Examples for logarithms
Logarithm expressions
Example 1.
A). x=10ac^2 (a>0,c>0).
Using properties 3.5 we calculate 
2.
3. 
4. 
Where 
.
Example 2. Find x if
Example 3. Let the value of logarithms be given
Calculate log(x) if
Basic properties of logarithms
Logarithms, like any numbers, can be added, subtracted and transformed in every way. But since logarithms are not exactly ordinary numbers, there are rules here, which are called main properties.
You definitely need to know these rules - without them not a single serious problem can be solved. logarithmic problem. In addition, there are very few of them - you can learn everything in one day. So let's get started.
Adding and subtracting logarithms
Consider two logarithms with the same bases: logax and logay. Then they can be added and subtracted, and:
- logax + logay = loga(x y);
- logax − logay = loga (x: y).
So, the sum of logarithms is equal to the logarithm of the product, and the difference is equal to the logarithm of the quotient. Note: key moment Here - identical grounds. If the reasons are different, these rules do not work!
These formulas will help you calculate a logarithmic expression even when its individual parts are not considered (see the lesson “What is a logarithm”). Take a look at the examples and see:
Since logarithms have the same bases, we use the sum formula:
log6 4 + log6 9 = log6 (4 9) = log6 36 = 2.
Task. Find the value of the expression: log2 48 − log2 3.
The bases are the same, we use the difference formula:
log2 48 − log2 3 = log2 (48: 3) = log2 16 = 4.
Task. Find the value of the expression: log3 135 − log3 5.
Again the bases are the same, so we have:
log3 135 − log3 5 = log3 (135: 5) = log3 27 = 3.
As you can see, the original expressions are made up of “bad” logarithms, which are not calculated separately. But after the transformations they turn out quite normal numbers. Many are built on this fact test papers. What about the controls? similar expressions in all seriousness (sometimes with virtually no changes) are offered on the Unified State Examination.
Extracting the exponent from the logarithm
It's easy to notice that last rule follows the first two. But it’s better to remember it anyway - in some cases it will significantly reduce the amount of calculations.
Of course, all these rules make sense if the ODZ of the logarithm is observed: a > 0, a ≠ 1, x > 0. And one more thing: learn to apply all formulas not only from left to right, but also vice versa, i.e. You can enter the numbers before the logarithm sign into the logarithm itself. This is what is most often required.
Task. Find the value of the expression: log7 496.
Let's get rid of the degree in the argument using the first formula:
log7 496 = 6 log7 49 = 6 2 = 12
Task. Find the meaning of the expression:
Note that the denominator contains a logarithm, the base and argument of which are exact powers: 16 = 24; 49 = 72. We have:
I think to last example clarification required. Where have logarithms gone? Until the very last moment we work only with the denominator.
Logarithm formulas. Logarithms examples solutions.
We presented the base and argument of the logarithm standing there in the form of powers and took out the exponents - we got a “three-story” fraction.
Now let's look at the main fraction. The numerator and denominator contain the same number: log2 7. Since log2 7 ≠ 0, we can reduce the fraction - 2/4 will remain in the denominator. According to the rules of arithmetic, the four can be transferred to the numerator, which is what was done. The result was the answer: 2.
Transition to a new foundation
Speaking about the rules for adding and subtracting logarithms, I specifically emphasized that they only work with the same bases. What if the reasons are different? What if they are not exact powers of the same number?
Formulas for transition to a new foundation come to the rescue. Let us formulate them in the form of a theorem:
Let the logarithm logax be given. Then for any number c such that c > 0 and c ≠ 1, the equality is true:
In particular, if we set c = x, we get:
From the second formula it follows that the base and argument of the logarithm can be swapped, but in this case the entire expression is “turned over”, i.e. the logarithm appears in the denominator.
These formulas are rarely found in conventional numerical expressions. It is possible to evaluate how convenient they are only when solving logarithmic equations and inequalities.
However, there are problems that cannot be solved at all except by moving to a new foundation. Let's look at a couple of these:
Task. Find the value of the expression: log5 16 log2 25.
Note that the arguments of both logarithms contain exact powers. Let's take out the indicators: log5 16 = log5 24 = 4log5 2; log2 25 = log2 52 = 2log2 5;
Now let’s “reverse” the second logarithm:
Since the product does not change when rearranging factors, we calmly multiplied four and two, and then dealt with logarithms.
Task. Find the value of the expression: log9 100 lg 3.
The base and argument of the first logarithm are exact powers. Let's write this down and get rid of the indicators:
Now let's get rid of the decimal logarithm by moving to a new base:
Basic logarithmic identity
Often in the solution process it is necessary to represent a number as a logarithm to a given base. In this case, the following formulas will help us:
In the first case, the number n becomes the exponent in the argument. The number n can be absolutely anything, because it is just a logarithm value.
The second formula is actually a paraphrased definition. That's what it's called: .
In fact, what happens if the number b is raised to such a power that the number b to this power gives the number a? That's right: the result is the same number a. Read this paragraph carefully again - many people get stuck on it.
Like the formulas for transition to a new base, the main logarithmic identity sometimes it is the only possible solution.
Task. Find the meaning of the expression:
Note that log25 64 = log5 8 - simply took the square from the base and argument of the logarithm. Considering the rules for multiplying powers with the same basis, we get:
If anyone doesn’t know, this was a real task from the Unified State Exam :)
Logarithmic unit and logarithmic zero
In conclusion, I will give two identities that can hardly be called properties - rather, they are consequences of the definition of the logarithm. They constantly appear in problems and, surprisingly, create problems even for “advanced” students.
- logaa = 1 is. Remember once and for all: the logarithm to any base a of that base itself equal to one.
- loga 1 = 0 is. The base a can be anything, but if the argument contains one - logarithm equal to zero! Because a0 = 1 is a direct consequence of the definition.
That's all the properties. Be sure to practice putting them into practice! Download the cheat sheet at the beginning of the lesson, print it out, and solve the problems.
See also:
The logarithm of b to base a denotes the expression. To calculate the logarithm means to find a power x () at which the equality is satisfied
Basic properties of the logarithm
It is necessary to know the above properties, since almost all problems and examples related to logarithms are solved on their basis. The rest of the exotic properties can be derived through mathematical manipulations with these formulas
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
When calculating the formula for the sum and difference of logarithms (3.4) you come across quite often. The rest are somewhat complex, but in a number of tasks they are indispensable for simplifying complex expressions and calculating their values.
Common cases of logarithms
Some of the common logarithms are those in which the base is even ten, exponential or two.
The logarithm to base ten is usually called the decimal logarithm and is simply denoted by lg(x).
It is clear from the recording that the basics are not written in the recording. For example
Natural logarithm is a logarithm with an exponent as its basis (denoted by ln(x)).
The exponent is 2.718281828…. To remember the exponent, you can study the rule: the exponent is equal to 2.7 and twice the year of birth of Leo Nikolaevich Tolstoy. Knowing this rule, you will know both the exact value of the exponent and the date of birth of Leo Tolstoy.
And another important logarithm to base two is denoted by
The derivative of the logarithm of a function is equal to one divided by the variable
Integral or antiderivative logarithm determined by dependency
The given material is enough for you to solve a wide class of problems related to logarithms and logarithms. To help you understand the material, I will give just a few common examples from school curriculum and universities.
Examples for logarithms
Logarithm expressions
Example 1.
A). x=10ac^2 (a>0,c>0).
Using properties 3.5 we calculate
2.
By the property of difference of logarithms we have
3.
Using properties 3.5 we find
4. Where .
By the look complex expression using a number of rules is simplified to form
Finding logarithm values
Example 2. Find x if
Solution. For calculation, we apply to the last term 5 and 13 properties
We put it on record and mourn
Since the bases are equal, we equate the expressions
Logarithms. First level.
Let the value of logarithms be given
Calculate log(x) if
Solution: Let's take a logarithm of the variable to write the logarithm through the sum of its terms
This is just the beginning of our acquaintance with logarithms and their properties. Practice calculations, enrich your practical skills - you will soon need the knowledge you gain to solve logarithmic equations. Having studied the basic methods for solving such equations, we will expand your knowledge for another no less important topic- logarithmic inequalities...
Basic properties of logarithms
Logarithms, like any numbers, can be added, subtracted and transformed in every way. But since logarithms are not exactly ordinary numbers, there are rules here, which are called main properties.
You definitely need to know these rules - without them, not a single serious logarithmic problem can be solved. In addition, there are very few of them - you can learn everything in one day. So let's get started.
Adding and subtracting logarithms
Consider two logarithms with the same bases: logax and logay. Then they can be added and subtracted, and:
- logax + logay = loga(x y);
- logax − logay = loga (x: y).
So, the sum of logarithms is equal to the logarithm of the product, and the difference is equal to the logarithm of the quotient. Please note: the key point here is identical grounds. If the reasons are different, these rules do not work!
These formulas will help you calculate a logarithmic expression even when its individual parts are not considered (see the lesson “What is a logarithm”). Take a look at the examples and see:
Task. Find the value of the expression: log6 4 + log6 9.
Since logarithms have the same bases, we use the sum formula:
log6 4 + log6 9 = log6 (4 9) = log6 36 = 2.
Task. Find the value of the expression: log2 48 − log2 3.
The bases are the same, we use the difference formula:
log2 48 − log2 3 = log2 (48: 3) = log2 16 = 4.
Task. Find the value of the expression: log3 135 − log3 5.
Again the bases are the same, so we have:
log3 135 − log3 5 = log3 (135: 5) = log3 27 = 3.
As you can see, the original expressions are made up of “bad” logarithms, which are not calculated separately. But after the transformations, completely normal numbers are obtained. Many tests are based on this fact. Yes, test-like expressions are offered in all seriousness (sometimes with virtually no changes) on the Unified State Examination.
Extracting the exponent from the logarithm
Now let's complicate the task a little. What if the base or argument of a logarithm is a power? Then the exponent of this degree can be taken out of the sign of the logarithm according to the following rules:
It is easy to see that the last rule follows the first two. But it’s better to remember it anyway - in some cases it will significantly reduce the amount of calculations.
Of course, all these rules make sense if the ODZ of the logarithm is observed: a > 0, a ≠ 1, x > 0. And one more thing: learn to apply all formulas not only from left to right, but also vice versa, i.e. You can enter the numbers before the logarithm sign into the logarithm itself.
How to solve logarithms
This is what is most often required.
Task. Find the value of the expression: log7 496.
Let's get rid of the degree in the argument using the first formula:
log7 496 = 6 log7 49 = 6 2 = 12
Task. Find the meaning of the expression:
Note that the denominator contains a logarithm, the base and argument of which are exact powers: 16 = 24; 49 = 72. We have:
I think the last example requires some clarification. Where have logarithms gone? Until the very last moment we work only with the denominator. We presented the base and argument of the logarithm standing there in the form of powers and took out the exponents - we got a “three-story” fraction.
Now let's look at the main fraction. The numerator and denominator contain the same number: log2 7. Since log2 7 ≠ 0, we can reduce the fraction - 2/4 will remain in the denominator. According to the rules of arithmetic, the four can be transferred to the numerator, which is what was done. The result was the answer: 2.
Transition to a new foundation
Speaking about the rules for adding and subtracting logarithms, I specifically emphasized that they only work with the same bases. What if the reasons are different? What if they are not exact powers of the same number?
Formulas for transition to a new foundation come to the rescue. Let us formulate them in the form of a theorem:
Let the logarithm logax be given. Then for any number c such that c > 0 and c ≠ 1, the equality is true:
In particular, if we set c = x, we get:
From the second formula it follows that the base and argument of the logarithm can be swapped, but in this case the entire expression is “turned over”, i.e. the logarithm appears in the denominator.
These formulas are rarely found in ordinary numerical expressions. It is possible to evaluate how convenient they are only when solving logarithmic equations and inequalities.
However, there are problems that cannot be solved at all except by moving to a new foundation. Let's look at a couple of these:
Task. Find the value of the expression: log5 16 log2 25.
Note that the arguments of both logarithms contain exact powers. Let's take out the indicators: log5 16 = log5 24 = 4log5 2; log2 25 = log2 52 = 2log2 5;
Now let’s “reverse” the second logarithm:
Since the product does not change when rearranging factors, we calmly multiplied four and two, and then dealt with logarithms.
Task. Find the value of the expression: log9 100 lg 3.
The base and argument of the first logarithm are exact powers. Let's write this down and get rid of the indicators:
Now let's get rid of the decimal logarithm by moving to a new base:
Basic logarithmic identity
Often in the solution process it is necessary to represent a number as a logarithm to a given base. In this case, the following formulas will help us:
In the first case, the number n becomes the exponent in the argument. The number n can be absolutely anything, because it is just a logarithm value.
The second formula is actually a paraphrased definition. That's what it's called: .
In fact, what happens if the number b is raised to such a power that the number b to this power gives the number a? That's right: the result is the same number a. Read this paragraph carefully again - many people get stuck on it.
Like formulas for moving to a new base, the basic logarithmic identity is sometimes the only possible solution.
Task. Find the meaning of the expression:
Note that log25 64 = log5 8 - simply took the square from the base and argument of the logarithm. Taking into account the rules for multiplying powers with the same base, we get:
If anyone doesn’t know, this was a real task from the Unified State Exam :)
Logarithmic unit and logarithmic zero
In conclusion, I will give two identities that can hardly be called properties - rather, they are consequences of the definition of the logarithm. They constantly appear in problems and, surprisingly, create problems even for “advanced” students.
- logaa = 1 is. Remember once and for all: the logarithm to any base a of that base itself is equal to one.
- loga 1 = 0 is. The base a can be anything, but if the argument contains one, the logarithm is equal to zero! Because a0 = 1 is a direct consequence of the definition.
That's all the properties. Be sure to practice putting them into practice! Download the cheat sheet at the beginning of the lesson, print it out, and solve the problems.
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 is equal to one, that is, lg10=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, 
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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).
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Check whether there are negative numbers or one under the logarithm sign. This method applicable to expressions of the form log b (x) log b (a) (\displaystyle (\frac (\log _(b)(x))(\log _(b)(a)))). However, it is not suitable for some special cases:
- Logarithm negative number not determined on any basis (for example, log (− 3) (\displaystyle \log(-3)) or log 4 (− 5) (\displaystyle \log _(4)(-5))). In this case write "no solution".
- The logarithm of zero to any base is also undefined. If you get caught ln (0) (\displaystyle \ln(0)), write down "no solution".
- Logarithm of one to any base ( log (1) (\displaystyle \log(1))) is always zero, because x 0 = 1 (\displaystyle x^(0)=1) for all values x. Write 1 instead of this logarithm and do not use the method below.
- If logarithms have different reasons, For example l o g 3 (x) l o g 4 (a) (\displaystyle (\frac (log_(3)(x))(log_(4)(a)))), and are not reduced to integers, the value of the expression cannot be found manually.
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Convert the expression to one logarithm. If the expression is not one of the above special occasions, it can be represented as a single logarithm. Use for this the following formula: log b (x) log b (a) = log a (x) (\displaystyle (\frac (\log _(b)(x))(\log _(b)(a)))=\ log_(a)(x)).
- Example 1: Consider the expression log 16 log 2 (\displaystyle (\frac (\log (16))(\log (2)))).
First, let's represent the expression as a single logarithm using the above formula: log 16 log 2 = log 2 (16) (\displaystyle (\frac (\log (16))(\log (2)))=\log _(2)(16)). - This formula for “replacing the base” of a logarithm is derived from the basic properties of logarithms.
- Example 1: Consider the expression log 16 log 2 (\displaystyle (\frac (\log (16))(\log (2)))).
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If possible, evaluate the value of the expression manually. To find log a (x) (\displaystyle \log _(a)(x)), imagine the expression " a? = x (\displaystyle a^(?)=x)", that is, ask yourself next question: "To what power should we raise a, To obtain x?. Answering this question may require a calculator, but if you're lucky, you may be able to find it manually.
- Example 1 (continued): Rewrite as 2? = 16 (\displaystyle 2^(?)=16). You need to find what number should stand in place of the "?" sign. This can be done by trial and error:
2 2 = 2 ∗ 2 = 4 (\displaystyle 2^(2)=2*2=4)
2 3 = 4 ∗ 2 = 8 (\displaystyle 2^(3)=4*2=8)
2 4 = 8 ∗ 2 = 16 (\displaystyle 2^(4)=8*2=16)
So the number we are looking for is 4: log 2 (16) (\displaystyle \log _(2)(16)) = 4 .
- Example 1 (continued): Rewrite as 2? = 16 (\displaystyle 2^(?)=16). You need to find what number should stand in place of the "?" sign. This can be done by trial and error:
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Leave your answer in logarithmic form if you can't simplify it. Many logarithms are very difficult to calculate by hand. In this case, to get an accurate answer, you will need a calculator. However, if you solve a task in class, the teacher will most likely be satisfied with the answer in logarithmic form. The method discussed below is used to solve a more complex example:
- example 2: what is equal to log 3 (58) log 3 (7) (\displaystyle (\frac (\log _(3)(58))(\log _(3)(7))))?
- Let's convert this expression to one logarithm: log 3 (58) log 3 (7) = log 7 (58) (\displaystyle (\frac (\log _(3)(58))(\log _(3)(7)))=\ log_(7)(58)). Note that the base 3 common to both logarithms disappears; this is true for any reason.
- Let's rewrite the expression in the form 7? = 58 (\displaystyle 7^(?)=58) and let's try to find the value?:
7 2 = 7 ∗ 7 = 49 (\displaystyle 7^(2)=7*7=49)
7 3 = 49 ∗ 7 = 343 (\displaystyle 7^(3)=49*7=343)
Because 58 is between these two numbers, it is not expressed as a whole number. - We leave the answer in logarithmic form: log 7 (58) (\displaystyle \log _(7)(58)).