Log by base. Definition of the logarithm and its properties: theory and problem solving
(from Greek λόγος - “word”, “relation” and ἀριθμός - “number”) numbers b based on a(log α b) is called such a number c, And b= a c, that is, records log α b=c And b=ac are equivalent. The logarithm makes sense if a > 0, a ≠ 1, b > 0.
In other words logarithm numbers b based on A formulated as an 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 α b, is equivalent to solving the equation a x =b.
For example:
log 2 8 = 3 because 8 = 2 3 .
Let us emphasize that the indicated formulation of the logarithm makes it possible to immediately determine logarithm value, when the number under the logarithm sign acts as a certain power of the base. Indeed, 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 powers of a number.
Calculating the logarithm is called logarithm. Logarithm is the mathematical operation of taking a logarithm. When taking logarithms, products of factors are transformed into sums of terms.
Potentiation is the inverse mathematical operation of logarithm. During potentiation, a given base is raised to the degree of expression over which potentiation is performed. In this case, the sums of terms are transformed into a product of factors.
Quite often, real logarithms are used with bases 2 (binary), Euler's number e ≈ 2.718 (natural logarithm) and 10 (decimal).
On at this stage it is advisable to consider logarithm samples log 7 2 , ln √ 5, lg0.0001.
And the entries lg(-3), log -3 3.2, log -1 -4.3 do not make sense, since in the first of them a negative number is placed under the logarithm sign, in the second - a negative number in the base, and in the third - both a negative number under the logarithm sign and a unit in the base.
Conditions for determining the logarithm.
It is worth considering separately the conditions a > 0, a ≠ 1, b > 0.under which we get definition of logarithm. Let's consider why these restrictions were taken. An equality of the form x = log α will help us with this b, called the basic logarithmic identity, which directly follows from the definition of logarithm given above.
Let's take the condition a≠1. Since one to any power is equal to one, then the equality x=log α b can only exist when b=1, but log 1 1 will be any real number. To eliminate this ambiguity, we take a≠1.
Let us prove the necessity of the condition a>0. At a=0 according to the formulation of the logarithm can exist only when b=0. And accordingly then log 0 0 can be any non-zero real number, since zero to any non-zero power is zero. This ambiguity can be eliminated by the condition a≠0. And when a<0 we would have to reject the analysis of rational and irrational values of the logarithm, since a degree with a rational and irrational exponent is defined only for non-negative bases. It is for this reason that the condition is stipulated a>0.
And the last condition b>0 follows from inequality a>0, since x=log α b, and the value of the degree with a positive base a always positive.
Features of logarithms.
Logarithms characterized by distinctive features, which led to their widespread use to significantly facilitate painstaking calculations. When moving “into the world of logarithms,” multiplication is transformed into a much easier addition, division is transformed into subtraction, and exponentiation and root extraction are transformed, respectively, into multiplication and division by the exponent.
The formulation of logarithms and a table of their values (for trigonometric functions) was first published in 1614 by the Scottish mathematician John Napier. Logarithmic tables, enlarged and detailed by other scientists, were widely used in scientific and engineering calculations, and remained relevant until the use of electronic calculators and computers.
Definition of logarithm
The logarithm of b to base a is the exponent to which a must be raised to get b.
Number e in mathematics it is customary to denote the limit to which an expression strives
Number e is irrational number- a number incommensurable with one, it cannot be accurately expressed as either an integer or a fraction rational number.
Letter e- first letter of a Latin word exponere- to show off, hence the name in mathematics exponential- exponential function.
Number e widely used in mathematics, and in all sciences that in one way or another use mathematical calculations for their needs.
Logarithms. Properties of logarithms
Definition: The logarithm of a positive number b to its base is the exponent c to which the number a must be raised to obtain the number b.
Basic logarithmic identity:
7) Formula for moving to a new base:
lna = log e a, e ≈ 2.718…
Problems and tests on the topic “Logarithms. Properties of logarithms"
- Logarithms - Important topics for reviewing the Unified State Examination in mathematics
To successfully complete tasks on this topic, you must know the definition of a logarithm, the properties of logarithms, the basic logarithmic identity, the definitions of decimal and natural logarithms. The main types of problems on this topic are problems involving the calculation and transformation of logarithmic expressions. Let's consider their solution using the following examples.
Solution: Using the properties of logarithms, we get
Solution: Using the properties of degrees, we get
1) (2 2) log 2 5 =(2 log 2 5) 2 =5 2 =25
Properties of logarithms, formulations and proofs.
Logarithms have a number of characteristic properties. In this article we will look at the main properties of logarithms. Here we will give their formulations, write down the properties of logarithms in the form of formulas, show examples of their application, and also provide proof of the properties of logarithms.
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Basic properties of logarithms, formulas
For ease of remembering and use, let’s imagine basic properties of logarithms in the form of a list of formulas. In the next paragraph we will give their formulations, evidence, examples of use and necessary explanations.
and the property of the logarithm of the product of n positive numbers: log a (x 1 · x 2 ·…·x n)= log a x 1 +log a x 2 +…+log a x n , a>0 , a≠1 , x 1 >0, x 2 >0, …, x n >0 .
, where a>0, a≠1, x>0, y>0.
, where a>0, a≠1, n – natural number, greater than one, b>0.
, a>0 , a≠1 , b>0 , b≠1 .
, a>0 , a≠1 , b>0 , p and q are real numbers, q≠0 , in particular for b=a we have 
.Formulations and proofs of properties
We proceed to the formulation and proof of the written properties of logarithms. All properties of logarithms are proved on the basis of the definition of the logarithm and the basic principle that follows from it logarithmic identity, as well as properties of degree.
Let's start with properties of the logarithm of one. Its formulation is as follows: the logarithm of unity is equal to zero, that is, log a 1=0 for any a>0, a≠1. The proof is not difficult: since a 0 =1 for any a satisfying the above conditions a>0 and a≠1, then the equality log a 1=0 to be proved follows immediately from the definition of the logarithm.
Let us give examples of the application of the considered property: log 3 1=0, log1=0 and .
Let's move on to the next property: the logarithm of a number equal to the base is equal to one, that is, log a a=1 for a>0, a≠1. Indeed, since a 1 =a for any a, then by definition of the logarithm log a a=1.
Examples of using this property of logarithms are the equalities log 5 5=1, log 5.6 5.6 and lne=1.
The logarithm of a power of a number equal to the base of the logarithm is equal to the exponent. This property of the logarithm corresponds to a formula of the form log a a p =p, where a>0, a≠1 and p – any real number. This property follows directly from the definition of the logarithm. Note that it allows you to immediately indicate the value of the logarithm, if it is possible to represent the number under the logarithm sign as a power of the base; we will talk more about this in the article calculating logarithms.
For example, log 2 2 7 =7, log10 -4 =-4 and 
.
Logarithm of the product of two positive numbers x and y equal to the product logarithms of these numbers: log a (x y)=log a x+log a y, a>0 , a≠1 . Let us prove the property of the logarithm of a product. Due to the properties of the degree a log a x+log a y =a log a x ·a log a y, and since by the main logarithmic identity a log a x =x and a log a y =y, then a log a x ·a log a y =x· y. Thus, a log a x+log a y =x·y, from which, by the definition of a logarithm, the equality being proved follows.
Let's show examples of using the property of the logarithm of a product: log 5 (2 3)=log 5 2+log 5 3 and 
.
The property of the logarithm of a product can be generalized to the product of a finite number n of positive numbers x 1 , x 2 , …, x n as log a (x 1 · x 2 ·…·x n)= log a x 1 +log a x 2 +…+log a x n. This equality can be proven without problems using the method of mathematical induction.
For example, the natural logarithm of the product can be replaced by the sum of three natural logarithms of the numbers 4, e, and.
Logarithm of the quotient of two positive numbers x and y is equal to the difference between the logarithms of these numbers. The property of the logarithm of a quotient corresponds to a formula of the form , where a>0, a≠1, x and y are some positive numbers. The validity of this formula is proven as well as the formula for the logarithm of a product: since 
, then by definition of the logarithm .
Here is an example of using this property of the logarithm: 
.
Let's move on to property of the logarithm of the power. The logarithm of a degree is equal to the product of the exponent and the logarithm of the modulus of the base of this degree. Let us write this property of the logarithm of a power as a formula: log a b p =p·log a |b|, where a>0, a≠1, b and p are numbers such that the degree b p makes sense and b p >0.
First we prove this property for positive b. The basic logarithmic identity allows us to represent the number b as a log a b , then b p =(a log a b) p , and the resulting expression, due to the property of power, is equal to a p·log a b . So we come to the equality b p =a p·log a b, from which, by the definition of a logarithm, we conclude that log a b p =p·log a b.
It remains to prove this property for negative b. Here we note that the expression log a b p for negative b makes sense only for even exponents p (since the value of the degree b p must be greater than zero, otherwise the logarithm will not make sense), and in this case b p =|b| p. Then b p =|b| p =(a log a |b|) p =a p·log a |b| , from where log a b p =p·log a |b| .
For example, 
and ln(-3) 4 =4·ln|-3|=4·ln3 .
It follows from the previous property property of the logarithm from the root: the logarithm of the nth root is equal to the product of the fraction 1/n by the logarithm of the radical expression, that is, where a>0, a≠1, n is a natural number greater than one, b>0.
The proof is based on the equality (see definition of exponent with a fractional exponent), which is valid for any positive b, and the property of the logarithm of the exponent: 
.
Here is an example of using this property: 
.
Now let's prove formula for moving to a new logarithm base kind 
. To do this, it is enough to prove the validity of the equality log c b=log a b·log c a. The basic logarithmic identity allows us to represent the number b as a log a b , then log c b=log c a log a b . It remains to use the property of the logarithm of the degree: log c a log a b =log a b·log c a . This proves the equality log c b=log a b·log c a, which means the formula for transition to a new base of the logarithm is also proven .
Let's show a couple of examples of using this property of logarithms: and 
.
The formula for moving to a new base allows you to move on to working with logarithms that have a “convenient” base. For example, it can be used to change to natural or decimal logarithms so that you can calculate the value of a logarithm from a table of logarithms. The formula for moving to a new logarithm base also allows, in some cases, to find the value of a given logarithm when the values of some logarithms with other bases are known.
Used frequently special case formulas for transition to a new base of the logarithm with c=b of the form. This shows that log a b and log b a are mutually inverse numbers. Eg, 
.
The formula is also often used, which is convenient for finding the values of logarithms. To confirm our words, we will show how it can be used to calculate the value of a logarithm of the form . We have 
. To prove the formula, it is enough to use the formula for moving to a new base of the logarithm a: 
.
It remains to prove the properties of comparison of logarithms.
Let's use the opposite method. Suppose that for a 1 >1, a 2 >1 and a 1 2 and for 0 1, log a 1 b≤log a 2 b is true. Based on the properties of logarithms, these inequalities can be rewritten as 
And 
respectively, and from them it follows that log b a 1 ≤log b a 2 and log b a 1 ≥log b a 2, respectively. Then, by the properties of powers with on the same grounds the equalities b log b a 1 ≥b log b a 2 and b log b a 1 ≥b log b a 2 must be satisfied, that is, a 1 ≥a 2 . So we came to a contradiction to the condition a 1 2. This completes the proof.
Basic properties of logarithms
- Materials for the lesson
- Download all formulas
- log a x n = n · log a x ;
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:
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:
Task. Find the value of the expression: 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 they turn out quite normal numbers. 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 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: 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 a x be given. Then for any number c such that c > 0 and c ≠ 1, the equality is true:
[Caption for the picture]
In particular, if we set 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 ordinary 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 the decimal logarithm by 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:
- n = log a a n
-
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: the basic logarithmic identity.
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.
[Caption for the picture]
Note that log 25 64 = log 5 8 - we 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:
[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: the logarithm to any base a of that base itself is equal to one.
- log a 1 = 0 is logarithmic zero. The base a can be anything, but if the argument contains one, the logarithm is 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.
Logarithm. Properties of the logarithm (addition and subtraction).
Properties of the logarithm follow 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.
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:
As we see, sum of logarithms equals the logarithm of the product, and difference logarithms- logarithm of the quotient. Moreover, this is true if the numbers A, X And at positive and a ≠ 1.
It is important to note that the main aspect in these formulas are the same bases. If the grounds are different, these rules do not apply!
The rules for adding and subtracting logarithms with the same bases are read not only from left to right, but also vice versa. As a result, we have the theorems for the logarithm of the product and the logarithm of the quotient.
Logarithm of the product two positive numbers is equal to the sum of their logarithms ; rephrasing this theorem we get the following if the numbers A, x And at positive and a ≠ 1, That:
Logarithm of the quotient two positive numbers is equal to the difference between the logarithms of the dividend and the divisor. To put it another way, if the numbers A, X And at positive and a ≠ 1, That:
Let us apply the above theorems to solve examples:
If the numbers x And at are negative, then product logarithm formula becomes meaningless. Thus, it is forbidden to write:
since the expressions log 2 (-8) and log 2 (-4) are not defined at all (logarithmic function at= log 2 X defined only for positive argument values X).
Product theorem applicable not only for two, but also for an unlimited number of factors. This means that for every natural k and any positive numbers x 1 , x 2 , . . . ,x n there is an identity:
From logarithm quotient theorem One more property of the logarithm can be obtained. It is common knowledge that log a 1= 0, therefore
This means there is an equality:
Logarithms of two reciprocal numbers for the same reason will differ from each other solely by sign. So:
Logarithm. Properties of logarithms
Logarithm. Properties of logarithms
Let's consider equality. Let us know the values of and and we want to find the value of .
That is, we are looking for the exponent by which we need to cock it to get .
Let
a variable can take on any real value, then the following restrictions are imposed on the variables: o" title="a>o"/> , 1″ title=»a1″/>, 0″ title=»b>0″/>If we know the values of and , and we are faced with the task of finding the unknown, then for this purpose a mathematical operation is introduced, which is called logarithm.
To find the value we take logarithm of a number By basis :
The logarithm of a number to its base is the exponent to which it must be raised to get .
That is basic logarithmic identity:
o» title=»a>o»/> , 1″ title=»a1″/>, 0″ title=»b>0″/>
is essentially a mathematical notation definitions of logarithm.
The mathematical operation of logarithm is the inverse of the operation of exponentiation, so properties of logarithms are closely related to the properties of degree.
Let's list the main properties of logarithms:
(o" title="a>o"/> , 1″ title=»a1″/>, 0″ title=»b>0″/>, 0,
d>0″/>, 1″ title=”d1″/>
4.
5.
The following group of properties allows you to represent the exponent of an expression under the sign of the logarithm, or standing at the base of the logarithm in the form of a coefficient in front of the sign of the logarithm:
6.
7.
8.
9.
The next group of formulas allows you to move from a logarithm with a given base to a logarithm with an arbitrary base, and is called formulas for transition to a new base:
10.
12. (corollary from property 11)
The following three properties are not well known, but they are often used when solving logarithmic equations, or when simplifying expressions containing logarithms:
13.
14.
15.
Special cases:
— decimal logarithm
— natural logarithmWhen simplifying expressions containing logarithms, a general approach is used:
1. Introducing decimals in the form of ordinary ones.
2. Mixed numbers represented as improper fractions.
3. We decompose the numbers at the base of the logarithm and under the sign of the logarithm into simple factors.
4. We try to reduce all logarithms to the same base.
5. Apply the properties of logarithms.
Let's look at examples of simplifying expressions containing logarithms.
Example 1.
Calculate:
Let's simplify all exponents: our task is to reduce them to logarithms, the base of which is the same number as the base of the exponent.
==(by property 7)=(by property 6) =
Let's substitute the indicators that we got into the original expression. We get:
Answer: 5.25
Example 2. Calculate:
Let's reduce all logarithms to base 6 (in this case, the logarithms from the denominator of the fraction will “migrate” to the numerator):
Let's decompose the numbers under the logarithm sign into simple factors:
Let's apply properties 4 and 6:
Let's introduce the replacement
We get:
Answer: 1
Logarithm . Basic logarithmic identity.
Properties of logarithms. Decimal logarithm. Natural logarithm.
Logarithm positive number N to base (b > 0, b 1) is the exponent x to which b must be raised to get N .
This entry is equivalent to the following: b x = N .
Examples: log 3 81 = 4, since 3 4 = 81;
log 1/3 27 = – 3, since (1/3) - 3 = 3 3 = 27.
The above definition of logarithm can be written as an identity:
Basic properties of logarithms.
2) log 1 = 0, since b 0 = 1 .
3) The logarithm of the product is equal to the sum of the logarithms of the factors:
4) The logarithm of the quotient is equal to the difference between the logarithms of the dividend and the divisor:
5) The logarithm of a power is equal to the product of the exponent and the logarithm of its base:
The consequence of this property is the following: logarithm of the root equal to the logarithm radical number divided by the power of the root:
6) If the base of the logarithm is a degree, then the value the inverse of the exponent can be taken out as a log rhyme:
The last two properties can be combined into one:
7) Transition modulus formula (i.e. transition from one logarithm base to another base):
In the special case when N=a we have:
Decimal logarithm called base logarithm 10. It is denoted lg, i.e. log 10 N= log N. Logarithms of numbers 10, 100, 1000, . p are 1, 2, 3, …, respectively, i.e. have so many positive
units, how many zeros are there in a logarithmic number after one. Logarithms of numbers 0.1, 0.01, 0.001, . p are respectively –1, –2, –3, …, i.e. have as many negative ones as there are zeros in the logarithmic number before one (including zero integers). The logarithms of other numbers have a fractional part called mantissa. The integer part of a logarithm is called characteristic. For practical use, decimal logarithms are most convenient.
Natural logarithm called base logarithm e. It is denoted by ln, i.e. log e N= log N. Number e is irrational, its approximate value is 2.718281828. It is the limit to which the number tends (1 + 1 / n) n with unlimited increase n(cm. first wonderful limit on the "Limits" page number sequences»).
Strange as it may seem, natural logarithms turned out to be very convenient when carrying out various types of operations related to the analysis of functions. Calculating logarithms to base e carried out much faster than for any other reason.
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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 in given point y"(1)=8*e^0=8
Video on the topic
Learn the table of elementary derivatives. This will significantly save time.
Sources:
- derivative of a constant
So, what is the difference between rational equation from the rational? If the unknown variable is under the sign square root, then the equation is considered irrational.
Instructions
The main method for solving such equations is the method of constructing both sides equations into a square. However. this is natural, the first thing you need to do is get rid of the sign. This method is not technically difficult, but sometimes it can lead to trouble. For example, the equation is v(2x-5)=v(4x-7). By squaring both sides you get 2x-5=4x-7. Solving such an equation is not difficult; x=1. But the number 1 will not be given equations. Why? Substitute one into the equation instead of the value of x. And the right and left sides will contain expressions that do not make sense, that is. This value is not valid for a square root. Therefore, 1 is an extraneous root, and therefore 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, in right side and then use the squaring method. solve the resulting rational equation and roots. But also another, more elegant one. Enter a new variable; vх=y. Accordingly, you will receive an equation of the form 2y2+y-3=0. That is, the usual quadratic equation. Find its roots; y1=1 and y2=-3/2. Next, solve two equations vх=1; vх=-3/2. The second equation has no roots; from the first we find that x=1. Don't forget to check the roots.
Solving identities is quite simple. To do this you need to do identity transformations until the goal is achieved. Thus, with the help of simple arithmetic operations, the task at hand will be solved.
You will need
- - paper;
- - pen.
Instructions
The simplest of such transformations are algebraic abbreviated multiplications (such as the square of the sum (difference), difference of squares, sum (difference), cube of the sum (difference)). In addition, there are many 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 from a textbook on 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 an 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 is a trigonometric function whose argument is a polynomial, then try using the change of variables method. In order to do this, replace the polynomial in the argument of the integrand with some new variable. Based on the relationship between the new and old variables, determine the new limits of integration. By differentiating this expression, find the new differential in . So you will get the new kind of the previous integral, close to or even corresponding to any tabular one.Solving integrals of the second kind
If the integral is an integral of the second kind, a vector form of the integrand, then you will need to use the rules for the transition from these integrals to scalar ones. One such rule is the Ostrogradsky-Gauss relation. This law allows 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.
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 the initially specified values of 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 is equal to the exponent. 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 there is a sufficiently large natural number under the logarithm sign, it does not hurt to factor 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 one 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 the 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 the next paragraph we will show how this is 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 logarithm table. When working in the decimal number system, 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 using a specific example - it’s clearer this 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). The third digit of the number 1.256 (digit 5) is found 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 logarithm table at the intersection of the marked row and marked columns (these numbers are highlighted in 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 of the resulting number, that is, we take log102.76332≈lg1.028·10 2. Now we apply the properties of the logarithm: lg1.028·10 2 =lg1.028+lg10 2 =lg1.028+2. Finally, we find the value of the logarithm lg1.028 from the table of decimal logarithms lg1.028≈0.0086+0.0034=0.012. As a result, the entire process of calculating the logarithm looks like this: log102.76332=log1.0276332 10 2 ≈lg1.028 10 2 = log1.028+lg10 2 =log1.028+2≈0.012+2=2.012.
In conclusion, it is worth noting that using a table of decimal logarithms you can calculate the approximate value of any logarithm. To do this, it is enough to use the transition formula to go to decimal logarithms, find their values in the table, and perform the remaining calculations.
For example, let's calculate log 2 3 . According to the formula for transition to a new base of the logarithm, we have . From the table of decimal logarithms we find log3≈0.4771 and log2≈0.3010. Thus, 
.
Bibliography.
- Kolmogorov A.N., Abramov A.M., Dudnitsyn Yu.P. and others. Algebra and the beginnings of analysis: Textbook for grades 10 - 11 of general education institutions.
- Gusev V.A., Mordkovich A.G. Mathematics (a manual for those entering technical schools).
The concept of logarithm and the basic logarithmic identity
The concept of logarithm and the basic logarithmic identity are closely related, because definition of logarithm in mathematical notation and is .
The basic logarithmic identity follows from the definition of the logarithm:
Definition 1
Logarithm they call the exponent $n$, when raised to which the numbers $a$ get the number $b$.
Note 1
Exponential equation$a^n=b$ for $a > 0$, $a \ne 1$ has no solutions for non-positive $b$ and has a single root for positive $b$. This root is called logarithm of the number $b$ to base $a$ and write down:
$a^(\log_(a) b)=b$.
Definition 2
Expression
$a^(\log_(a) b)=b$
called basic logarithmic identity provided that $a,b > 0$, $a \ne 1$.
Example 1
$17^(\log_(17) 6)=6$;
$e^(\ln13) =13$;
$10^(\lg23)=23$.
Basic logarithmic identity
Main the logarithmic identity is called because it is used almost always when working with logarithms. In addition, with its help the basic properties of logarithms are substantiated.
Example 2
$7^5=16,807$, therefore $\log_(7)16,807=5$.
$3^(-5)=\frac(1)(243)$, therefore $\log_(3)\frac(1)(243)=-5$.
$11^0=1$, therefore $\log_(11)1=0$.
Let's consider a consequence of the basic logarithmic identity:
Definition 3
If two logarithms with the same bases are equal, then the logarithmic expressions are equal:
if $\log_(a)b=\log_(a)c$, then $b=c$.
Let's consider restrictions, which are used for the logarithmic identity:
Because when raising unity to any power we always get one, and the equality $x=\log_(a)b$ exists only for $b=1$, then $\log_(1)1$ will be any real number. To avoid this ambiguity, take $a \ne 1$.
The logarithm for $a=0$, according to definition, can exist only for $b=0$. Because When we raise zero to any power we always get zero, then $\log_(0)0$ can be any real number. To avoid this ambiguity, take $a \ne 0$. For $a rational and irrational logarithm values, because a degree with a rational and irrational exponent can only be calculated for positive bases. To prevent this situation, take $a > 0$.
$b > 0$ follows from the condition $a > 0$, since $x=\log_(a)b$, and the power of a positive number a will always be positive.
The basic logarithmic identity is often used to simplify logarithmic expressions.
Example 3
Calculate $81^(\log_(9) 7)$.
Solution.
In order to use the basic logarithmic identity, it is necessary that the base of the logarithm and the powers be the same. Let us write the base of the degree in the form:
Now we can write:
$81^(\log_(9)7)=(9^2)^(\log_(9)7)=$
Let's use the power property:
$=9^(2 \cdot \log_(9)7)=9^(\log_(9)7) \cdot 9^(\log_(9)7)=$
the basic logarithmic identity can now be applied to each factor:
$=7 \cdot 7=49$.
Note 2
To apply the basic logarithmic identity, you can also resort to replacing the base of the logarithm with the expression that appears under the logarithm sign, and vice versa.
Example 4
Calculate $7^(\frac(1)(\log_(11) 7))$.
Solution.
$7^(\frac(1)(\log_(11) 7))=7^(\log_(7) 11)=11$.
Answer: $11$.
Example 5
Calculate $7^(\frac(3)(\log_(11) 7))$.