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: log a x and log a y. Then they can be added and subtracted, and:
- log a x+ log a y=log a (x · y);
- log a x− log a y=log a (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 lesson “What is a logarithm”). Take a look at the examples and see:
Log 6 4 + log 6 9.
Since logarithms have the same bases, we use the sum formula:
log 6 4 + log 6 9 = log 6 (4 9) = log 6 36 = 2.
Task. Find the value of the expression: log 2 48 − log 2 3.
The bases are the same, we use the difference formula:
log 2 48 − log 2 3 = log 2 (48: 3) = log 2 16 = 4.
Task. Find the value of the expression: log 3 135 − log 3 5.
Again the bases are the same, so we have:
log 3 135 − log 3 5 = log 3 (135: 5) = log 3 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
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'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: log 7 49 6 .
Let's get rid of the degree in the argument using the first formula:
log 7 49 6 = 6 log 7 49 = 6 2 = 12
Task. Find the meaning of the expression:
[Caption for the picture]
Note that the denominator contains a logarithm, the base and argument of which are exact powers: 16 = 2 4 ; 49 = 7 2. We have:
[Caption for the picture] I think to last example clarification required. 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: log 2 7. Since log 2 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 log be given a x. Then for any number c such that c> 0 and c≠ 1, the equality is true:
[Caption for the picture]
In particular, if we put c = x, we get:
[Caption for the picture]
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 by deciding 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: log 5 16 log 2 25.
Note that the arguments of both logarithms contain exact powers. Let's take out the indicators: log 5 16 = log 5 2 4 = 4log 5 2; log 2 25 = log 2 5 2 = 2log 2 5;
Now let’s “reverse” the second logarithm:
[Caption for the picture]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: log 9 100 lg 3.
The base and argument of the first logarithm are exact powers. Let's write this down and get rid of the indicators:
[Caption for the picture]Now let's get rid of decimal logarithm, moving to a new base:
[Caption for the picture]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 an indicator of the degree standing in the argument. Number n can be absolutely anything, because it’s just a logarithm value.
The second formula is actually a paraphrased definition. That’s what it’s called: the basic logarithmic identity.
In fact, what will happen if the number b raise to such a power that the number b to this power gives the number a? That's right: you get this 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:
[Caption for the picture]
Note that log 25 64 = log 5 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:
[Caption for the picture]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.
- log a a= 1 is a logarithmic unit. Remember once and for all: logarithm to any base a from this very base is equal to one.
- log a 1 = 0 is logarithmic zero. Base a can be anything, but if the argument contains one - logarithm equal to zero! Because a 0 = 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.
Instructions
Write 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 at a 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 this equation has no roots.
So, an irrational equation is solved using the method of squaring both its sides. 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, it is necessary to carry out identical transformations until the set goal is achieved. Thus, with the help of simple arithmetic operations, the problem posed 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 is equal to the square of the first plus twice 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, what a definite integral is. As is known, the solution definite integral there is a function whose derivative gives an integrand. This function is called antiderivative. Based on this principle, the main integrals are constructed.Determine by the type of the integrand which of the table integrals is suitable 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. Differentiation given expression find a 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 us to move from the rotor flux of a certain 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 of the upper limit into the 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.
Logarithmic equations. We continue to consider problems from Part B of the Unified State Examination in mathematics. We have already examined solutions to some equations in the articles “”, “”. In this article we will look at logarithmic equations. I’ll say right away that there will be no complex transformations when solving such equations on the Unified State Exam. They are simple.
It is enough to know and understand the basic logarithmic identity, to know the properties of the logarithm. Please note that after solving it, you MUST do a check - substitute the resulting value into the original equation and calculate, in the end you should get the correct equality.
Definition:
The logarithm of a number to base b is the exponent.to which b must be raised to obtain a.
For example:
Log 3 9 = 2, since 3 2 = 9
Properties of logarithms:
Special cases of logarithms:
Let's solve problems. In the first example we will do a check. In the future, check it yourself.
Find the root of the equation: log 3 (4–x) = 4
Since log b a = x b x = a, then
3 4 = 4 – x
x = 4 – 81
x = – 77
Examination:
log 3 (4–(–77)) = 4
log 3 81 = 4
3 4 = 81 Correct.
Answer: – 77
Decide for yourself:
Find the root of the equation: log 2 (4 – x) = 7
Find the root of the equation log 5(4 + x) = 2
We use the basic logarithmic identity.
Since log a b = x b x = a, then
5 2 = 4 + x
x =5 2 – 4
x = 21
Examination:
log 5 (4 + 21) = 2
log 5 25 = 2
5 2 = 25 Correct.
Answer: 21
Find the root of the equation log 3 (14 – x) = log 3 5.
The following property takes place, its meaning is as follows: if on the left and right sides of the equation we have logarithms with the same base, then we can equate the expressions under the signs of the logarithms.
14 – x = 5
x=9
Do a check.
Answer: 9
Decide for yourself:
Find the root of the equation log 5 (5 – x) = log 5 3.
Find the root of the equation: log 4 (x + 3) = log 4 (4x – 15).
If log c a = log c b, then a = b
x + 3 = 4x – 15
3x = 18
x=6
Do a check.
Answer: 6
Find the root of the equation log 1/8 (13 – x) = – 2.
(1/8) –2 = 13 – x
8 2 = 13 – x
x = 13 – 64
x = – 51
Do a check.
A small addition - the property is used here
degrees ().
Answer: – 51
Decide for yourself:
Find the root of the equation: log 1/7 (7 – x) = – 2
Find the root of the equation log 2 (4 – x) = 2 log 2 5.
Let's transform the right side. Let's use the property:
log a b m = m∙log a b
log 2 (4 – x) = log 2 5 2
If log c a = log c b, then a = b
4 – x = 5 2
4 – x = 25
x = – 21
Do a check.
Answer: – 21
Decide for yourself:
Find the root of the equation: log 5 (5 – x) = 2 log 5 3
Solve the equation log 5 (x 2 + 4x) = log 5 (x 2 + 11)
If log c a = log c b, then a = b
x 2 + 4x = x 2 + 11
4x = 11
x = 2.75
Do a check.
Answer: 2.75
Decide for yourself:
Find the root of the equation log 5 (x 2 + x) = log 5 (x 2 + 10).
Solve the equation log 2 (2 – x) = log 2 (2 – 3x) +1.
Required with right side equations obtain an expression of the form:
log 2 (......)
We represent 1 as a base 2 logarithm:
1 = log 2 2
log c (ab) = log c a + log c b
log 2 (2 – x) = log 2 (2 – 3x) + log 2 2
We get:
log 2 (2 – x) = log 2 2 (2 – 3x)
If log c a = log c b, then a = b, then
2 – x = 4 – 6x
5x = 2
x = 0.4
Do a check.
Answer: 0.4
Decide for yourself: Next you need to solve the quadratic equation. By the way,
the roots are 6 and – 4.
Root "–4" is not a solution, since the base of the logarithm must be greater than zero, and with "– 4" it is equal to " – 5". The solution is root 6.Do a check.
Answer: 6.
R eat on your own:
Solve the equation log x –5 49 = 2. If the equation has more than one root, answer with the smaller one.
As you have seen, no complicated transformations with logarithmic equationsNo. It is enough to know the properties of the logarithm and be able to apply them. IN Unified State Examination tasks When converting logarithmic expressions, more serious conversions are performed and more advanced solution skills are required. We will look at such examples, don’t miss them!I wish you success!!!
Sincerely, Alexander Krutitskikh.
P.S: I would be grateful if you tell me about the site on social networks.
Examples:
\(\log_(2)(x) = 32\)
\(\log_3x=\log_39\)
\(\log_3((x^2-3))=\log_3((2x))\)
\(\log_(x+1)((x^2+3x-7))=2\)
\(\lg^2((x+1))+10=11 \lg((x+1))\)
How to solve logarithmic equations:
When solving a logarithmic equation, you should strive to transform it to the form \(\log_a(f(x))=\log_a(g(x))\), and then make the transition to \(f(x)=g(x) \).
\(\log_a(f(x))=\log_a(g(x))\) \(⇒\) \(f(x)=g(x)\).
Example:\(\log_2(x-2)=3\)
|
Solution: |
ODZ: |
Very important! This transition can only be made if:
You have written for the original equation, and at the end you will check whether those found are included in the DL. If this is not done, extra roots may appear, which means a wrong decision.
The number (or expression) on the left and right is the same;
The logarithms on the left and right are “pure”, that is, there should be no multiplications, divisions, etc. – only single logarithms on either side of the equal sign.
For example:
Note that equations 3 and 4 can be easily solved by applying required properties logarithms.
Example . Solve the equation \(2\log_8x=\log_82.5+\log_810\)
Solution :
|
Let's write the ODZ: \(x>0\). |
||
|
\(2\log_8x=\log_82.5+\log_810\) ODZ: \(x>0\) |
On the left in front of the logarithm is the coefficient, on the right is the sum of the logarithms. This bothers us. Let's move the two to the exponent \(x\) according to the property: \(n \log_b(a)=\log_b(a^n)\). Let us represent the sum of logarithms as one logarithm according to the property: \(\log_ab+\log_ac=\log_a(bc)\) |
|
|
\(\log_8(x^2)=\log_825\) |
We reduced the equation to the form \(\log_a(f(x))=\log_a(g(x))\) and wrote down the ODZ, which means we can move to the form \(f(x)=g(x)\ ). |
|
|
Happened . We solve it and get the roots. |
||
|
\(x_1=5\) \(x_2=-5\) |
We check whether the roots are suitable for ODZ. To do this, in \(x>0\) instead of \(x\) we substitute \(5\) and \(-5\). This operation can be performed orally. |
|
|
\(5>0\), \(-5>0\) |
The first inequality is true, the second is not. This means that \(5\) is the root of the equation, but \(-5\) is not. We write down the answer. |
Answer : \(5\)
Example : Solve the equation \(\log^2_2(x)-3 \log_2(x)+2=0\)
Solution :
|
Let's write the ODZ: \(x>0\). |
||
|
\(\log^2_2(x)-3 \log_2(x)+2=0\) ODZ: \(x>0\) |
A typical equation solved using . Replace \(\log_2x\) with \(t\). |
|
|
\(t=\log_2x\) |
||
|
We received the usual one. We are looking for its roots. |
||
|
\(t_1=2\) \(t_2=1\) |
Making a reverse replacement |
|
|
\(\log_2(x)=2\) \(\log_2(x)=1\) |
We transform the right-hand sides, representing them as logarithms: \(2=2 \cdot 1=2 \log_22=\log_24\) and \(1=\log_22\) |
|
|
\(\log_2(x)=\log_24\) \(\log_2(x)=\log_22 \) |
Now our equations are \(\log_a(f(x))=\log_a(g(x))\), and we can transition to \(f(x)=g(x)\). |
|
|
\(x_1=4\) \(x_2=2\) |
We check the correspondence of the roots of the ODZ. To do this, substitute \(4\) and \(2\) into the inequality \(x>0\) instead of \(x\). |
|
|
\(4>0\) \(2>0\) |
Both inequalities are true. This means that both \(4\) and \(2\) are roots of the equation. |
Answer : \(4\); \(2\).
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 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:
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
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 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
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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 only a few common examples from the 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
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By the property of difference of logarithms we have
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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.