Logarithm 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 a positive number that is 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 using 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 range of acceptable values, and this is categorically unacceptable, since it can lead to a loss of solutions. A 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) |
Today we will talk about logarithmic formulas and we will give indicative solution examples.
They themselves imply solution patterns according to the basic properties of logarithms. Before applying logarithm formulas to solve, let us remind you of all the properties:
Now, based on these formulas (properties), we will show examples of solving logarithms.
Examples of solving logarithms based on formulas.
Logarithm a positive number b to base a (denoted by log a b) is an exponent to which a must be raised to get b, with b > 0, a > 0, and 1.
According to the definition, log a b = x, which is equivalent to a x = b, therefore log a a x = x.
Logarithms, examples:
log 2 8 = 3, because 2 3 = 8
log 7 49 = 2, because 7 2 = 49
log 5 1/5 = -1, because 5 -1 = 1/5
Decimal logarithm- this is an ordinary logarithm, the base of which is 10. It is denoted as lg.
log 10 100 = 2, because 10 2 = 100
Natural logarithm- also an ordinary logarithm, a logarithm, but with the base e (e = 2.71828... - an irrational number). Denoted as ln.
It is advisable to memorize the formulas or properties of logarithms, because we will need them later when solving logarithms, logarithmic equations and inequalities. Let's work through each formula again with examples.
- Basics logarithmic identity
a log a b = b8 2log 8 3 = (8 2log 8 3) 2 = 3 2 = 9
- Logarithm of the product equal to the sum logarithms
log a (bc) = log a b + log a clog 3 8.1 + log 3 10 = log 3 (8.1*10) = log 3 81 = 4
- Logarithm of the quotient equal to the difference logarithms
log a (b/c) = log a b - log a c9 log 5 50 /9 log 5 2 = 9 log 5 50- log 5 2 = 9 log 5 25 = 9 2 = 81
- Properties of the power of a logarithmic number and the base of the logarithm
Exponent of the logarithmic number log a b m = mlog a b
Exponent of the base of the logarithm log a n b =1/n*log a b
log a n b m = m/n*log a b,
if m = n, we get log a n b n = log a b
log 4 9 = log 2 2 3 2 = log 2 3
- Transition to a new foundation
log a b = log c b/log c a,if c = b, we get log b b = 1
then log a b = 1/log b a
log 0.8 3*log 3 1.25 = log 0.8 3*log 0.8 1.25/log 0.8 3 = log 0.8 1.25 = log 4/5 5/4 = -1
As you can see, the formulas for logarithms are not as complicated as they seem. Now, having looked at examples of solving logarithms, we can move on to logarithmic equations. We will look at examples of solving logarithmic equations in more detail in the article: "". Do not miss!
If you still have questions about the solution, write them in the comments to the article.
Note: we decided to get a different class of education and study abroad as an option.
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 the logarithm logax be given. 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 both the exact value of the exponent 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 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 on the same grounds: 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 logarithmic expression even when its individual parts are not counted (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, completely normal numbers are obtained. Many are built on this fact test papers. 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
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. 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.
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 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 direct consequence from 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
A natural logarithm is a logarithm whose base is an exponent (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
The integral or antiderivative logarithm is determined by the relationship
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 to another equally 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.
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.
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