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Instruction

Write down 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 the base, then the expression is written: ln b is the natural logarithm. It is understood that the result of any is the power to which the base number must be raised to get the number b.

When finding the sum of two functions, you just 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, from the product of the derivative of the dividend multiplied by the divisor function, to subtract the product of the derivative of the divisor multiplied by the divisor function, and divide all this by the divisor function squared. (u/v)" = (u"*v-v"*u)/v^2;

If a complex function is given, then it is necessary to multiply the derivative of the inner function and the derivative of the outer one. Let y=u(v(x)), then y"(x)=y"(u)*v"(x).

Using the 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 tasks for 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 the given point y"(1)=8*e^0=8

Related videos

Helpful advice

Learn the table of elementary derivatives. This will save a lot of time.

Sources:

• constant derivative

So what is the difference between an irrational equation and a rational one? If the unknown variable is under the square root sign, then the equation is considered irrational.

Instruction

The main method for solving such equations is the method of raising both parts equations into a square. However. this is natural, the first step is to get rid of the sign. Technically, this method is not difficult, but sometimes it can lead to trouble. For example, the equation v(2x-5)=v(4x-7). By squaring both sides, you get 2x-5=4x-7. Such an equation is not difficult to solve; x=1. But the number 1 will not be given equations. Why? Substitute the unit in the equation instead of the x value. And the right and left sides will contain expressions that do not make sense, that is. Such a value is not valid for a square root. Therefore, 1 is an extraneous root, and therefore this equation has no roots.

So, the irrational equation is solved using the method of squaring both of its parts. And having solved the equation, it is necessary to cut off extraneous roots. To do this, substitute the found roots in the original equation.

Consider another one.
2x+vx-3=0
Of course, this equation can be solved using the same equation as the previous one. Transfer 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 another, more elegant one. Enter a new variable; vx=y. Accordingly, you will get an equation like 2y2+y-3=0. That is the usual quadratic equation. Find its roots; y1=1 and y2=-3/2. Next, solve two equations vx=1; vx \u003d -3/2. The second equation has no roots, from the first we find that x=1. Do not forget about the need to check the roots.

Solving identities is quite easy. This requires making identical transformations until the goal is achieved. Thus, with the help of the simplest arithmetic operations, the task will be solved.

You will need

• - paper;
• - pen.

Instruction

The simplest such transformations are algebraic abbreviated multiplications (such as the square of the sum (difference), the difference of squares, the sum (difference), the cube of the sum (difference)). In addition, there are many trigonometric formulas that 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 and the second 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 solution

Repeat from a textbook on mathematical analysis or higher mathematics, which is a definite integral. As you know, the solution of a definite integral is a function whose derivative will give an integrand. This function is called antiderivative. According to this principle, the basic integrals are constructed.
Determine by the form 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 substitution method

If the integrand is a trigonometric function whose argument is some polynomial, then try using the change of variables method. To do this, replace the polynomial in the argument of the integrand with some new variable. Based on the ratio between the new and old variable, determine the new limits of integration. By differentiating this expression, find a new differential in . Thus, you will get a new form of the old integral, close or even corresponding to any tabular one.

Solution of integrals of the second kind

If the integral is an integral of the second kind, the vector form of the integrand, then you will need to use the rules for moving from these integrals to scalar ones. One such rule is the Ostrogradsky-Gauss ratio. This law makes it possible to pass from the rotor flow of some vector function to a triple integral over the divergence of a given vector field.

Substitution of limits of integration

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 receive some number. Next, subtract from the resulting number another number, the resulting lower limit to the antiderivative. If one of the integration limits is infinity, then when substituting it into the antiderivative function, it is necessary to go to the limit and find what the expression tends to.
If the integral is two-dimensional or three-dimensional, then you will have to represent the geometric limits of integration in order to understand how to calculate the integral. Indeed, in the case of, say, a three-dimensional integral, the limits of integration can be entire planes that limit the volume to be integrated.

So, we have powers of two. If you take the number from the bottom line, then you can easily find the power to which you have to raise a two to get this number. For example, to get 16, you need to raise two to the fourth power. And to get 64, you need to raise two to the sixth power. This can be seen from the table.

And now - in fact, the definition of the logarithm:

The logarithm to the base a of the argument x is the power to which the number a must be raised to get the number x .

Notation: log a x \u003d b, where a is the base, x is the argument, b is actually what the logarithm is equal to.

For example, 2 3 = 8 ⇒ log 2 8 = 3 (the base 2 logarithm of 8 is three because 2 3 = 8). Might as well log 2 64 = 6 because 2 6 = 64 .

The operation of finding the logarithm of a number to a given base is called the logarithm. So let's add a new row to our table:

2 1	2 2	2 3	2 4	2 5	2 6
2	4	8	16	32	64
log 2 2 = 1	log 2 4 = 2	log 2 8 = 3	log 2 16 = 4	log 2 32 = 5	log 2 64 = 6

Unfortunately, not all logarithms are considered so easily. For example, try to find log 2 5 . The number 5 is not in the table, but logic dictates that the logarithm will lie somewhere on the segment. Because 2 2< 5 < 2 3 , а чем больше степень двойки, тем больше получится число.

Such numbers are called irrational: the numbers after the decimal point can be written indefinitely, and they never repeat. If the logarithm turns out to be irrational, it is better to leave it like this: log 2 5 , log 3 8 , log 5 100 .

It is important to understand that the logarithm is an expression with two variables (base and argument). At first, many people confuse where the base is and where the argument is. To avoid annoying misunderstandings, just take a look at the picture:

Before us is nothing more than the definition of the logarithm. Remember: the logarithm is the power, to which you need to raise the base to get the argument. It is the base that is raised to a power - in the picture it is highlighted in red. It turns out that the base is always at the bottom! I tell this wonderful rule to my students at the very first lesson - and there is no confusion.

We figured out the definition - it remains to learn how to count logarithms, i.e. get rid of the "log" sign. To begin with, we note that two important facts follow from the definition:

1. The argument and base must always be greater than zero. This follows from the definition of the degree by a rational exponent, to which the definition of the logarithm is reduced.
2. The base must be different from unity, since a unit to any power is still a unit. Because of this, the question “to what power must one be raised to get two” is meaningless. There is no such degree!

Such restrictions are called valid range(ODZ). It turns out that the ODZ of the logarithm looks like this: log a x = b ⇒ x > 0 , a > 0 , a ≠ 1 .

Note that there are no restrictions on the number b (the value of the logarithm) is not imposed. For example, the logarithm may well be negative: log 2 0.5 \u003d -1, because 0.5 = 2 −1 .

However, now we are considering only numerical expressions, where it is not required to know the ODZ of the logarithm. All restrictions have already been taken into account by the compilers of the problems. But when logarithmic equations and inequalities come into play, the DHS requirements will become mandatory. Indeed, in the basis and argument there can be very strong constructions that do not necessarily correspond to the above restrictions.

Now consider the general scheme for calculating logarithms. It consists of three steps:

1. Express the base a and the argument x as a power with the smallest possible base greater than one. Along the way, it is better to get rid of decimal fractions;
2. Solve the equation for the variable b: x = a b ;
3. The resulting number b will be the answer.

That's all! If the logarithm turns out to be irrational, this will be seen already at the first step. The requirement that the base be greater than one is very relevant: this reduces the likelihood of error and greatly simplifies calculations. Similarly with decimal fractions: if you immediately convert them to ordinary ones, there will be many times less errors.

Let's see how this scheme works with specific examples:

Task. Calculate the logarithm: log 5 25

1. Let's represent the base and the argument as a power of five: 5 = 5 1 ; 25 = 52;

Let's make and solve the equation:
log 5 25 = b ⇒ (5 1) b = 5 2 ⇒ 5 b = 5 2 ⇒ b = 2 ;

2. Received an answer: 2.

Task. Calculate the logarithm:

Task. Calculate the logarithm: log 4 64

1. Let's represent the base and the argument as a power of two: 4 = 2 2 ; 64 = 26;
2. Let's make and solve the equation:
   log 4 64 = b ⇒ (2 2) b = 2 6 ⇒ 2 2b = 2 6 ⇒ 2b = 6 ⇒ b = 3 ;
3. Received an answer: 3.

Task. Calculate the logarithm: log 16 1

1. Let's represent the base and the argument as a power of two: 16 = 2 4 ; 1 = 20;
2. Let's make and solve the equation:
   log 16 1 = b ⇒ (2 4) b = 2 0 ⇒ 2 4b = 2 0 ⇒ 4b = 0 ⇒ b = 0 ;
3. Received a response: 0.

Task. Calculate the logarithm: log 7 14

1. Let's represent the base and the argument as a power of seven: 7 = 7 1 ; 14 is not represented as a power of seven, because 7 1< 14 < 7 2 ;
2. It follows from the previous paragraph that the logarithm is not considered;
3. The answer is no change: log 7 14.

A small note on the last example. How to make sure that a number is not an exact power of another number? Very simple - just decompose it into prime factors. If there are at least two distinct factors in the expansion, the number is not an exact power.

Task. Find out if the exact powers of the number are: 8; 48; 81; 35; fourteen .

8 \u003d 2 2 2 \u003d 2 3 - the exact degree, because there is only one multiplier;
48 = 6 8 = 3 2 2 2 2 = 3 2 4 is not an exact power because there are two factors: 3 and 2;
81 \u003d 9 9 \u003d 3 3 3 3 \u003d 3 4 - exact degree;
35 = 7 5 - again not an exact degree;
14 \u003d 7 2 - again not an exact degree;

Note also that the prime numbers themselves are always exact powers of themselves.

Decimal logarithm

Some logarithms are so common that they have a special name and designation.

The decimal logarithm of the x argument is the base 10 logarithm, i.e. the power to which you need to raise the number 10 to get the number x. Designation: lg x .

For example, log 10 = 1; log 100 = 2; lg 1000 = 3 - etc.

From now on, when a phrase like “Find lg 0.01” appears in the textbook, know that this is not a typo. This is the decimal logarithm. However, if you are not used to such a designation, you can always rewrite it:
log x = log 10 x

Everything that is true for ordinary logarithms is also true for decimals.

natural logarithm

There is another logarithm that has its own notation. In a sense, it is even more important than decimal. This is the natural logarithm.

The natural logarithm of x is the base e logarithm, i.e. the power to which the number e must be raised to obtain the number x. Designation: ln x .

Many will ask: what else is the number e? This is an irrational number, its exact value cannot be found and written down. Here are just the first numbers:
e = 2.718281828459...

We will not delve into what this number is and why it is needed. Just remember that e is the base of the natural logarithm:
ln x = log e x

Thus ln e = 1 ; log e 2 = 2 ; ln e 16 = 16 - etc. On the other hand, ln 2 is an irrational number. In general, the natural logarithm of any rational number is irrational. Except, of course, unity: ln 1 = 0.

For natural logarithms, all the rules that are true for ordinary logarithms are valid.

Logarithms, like any number, can be added, subtracted and converted in every possible way. But since logarithms are not quite ordinary numbers, there are rules here, which are called basic properties.

These rules must be known - no serious logarithmic problem can be solved without them. In addition, there are very few of them - everything can be learned in one day. So let's get started.

Addition and subtraction of logarithms

Consider two logarithms with the same base: log a x and log a y. Then they can be added and subtracted, and:

1. log a x+log a y= log a (x · y);
2. log a x−log a y= log a (x : y).

So, the sum of the logarithms is equal to the logarithm of the product, and the difference is the logarithm of the quotient. Please note: the key point here is - same grounds. If the bases are different, these rules do not work!

These formulas will help you calculate the 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:

log 6 4 + log 6 9.

Since the bases of logarithms are the same, 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 considered separately. But after transformations quite normal numbers turn out. Many tests are based on this fact. Yes, control - similar expressions in all seriousness (sometimes - with virtually no changes) are offered at the exam.

Removing the exponent from the logarithm

Now let's complicate the task a little. What if there is a degree in the base or argument of the logarithm? 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 their 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 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 sign of the logarithm 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 according to the first formula:
log 7 49 6 = 6 log 7 49 = 6 2 = 12

Task. Find the value of the expression:

[Figure caption]

Note that the denominator is a logarithm whose base and argument are exact powers: 16 = 2 4 ; 49 = 72. We have:

[Figure caption]

I think the last example needs clarification. Where have logarithms gone? Until the very last moment, we work only with the denominator. They presented the base and the argument of the logarithm standing there in the form of degrees and took out the indicators - they got a “three-story” fraction.

Now let's look at the main fraction. The numerator and denominator have 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 was done. The result is 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 bases are different? What if they are not exact powers of the same number?

Formulas for transition to a new base come to the rescue. We formulate them in the form of a theorem:

Let the logarithm log a x. Then for any number c such that c> 0 and c≠ 1, the equality is true:

[Figure caption]

In particular, if we put c = x, we get:

[Figure caption]

It follows from the second formula that it is possible to interchange the base and the argument of the logarithm, but in this case the whole expression is “turned over”, i.e. the logarithm is 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 tasks that cannot be solved at all except by moving to a new foundation. Let's consider a couple of these:

Task. Find the value of the expression: log 5 16 log 2 25.

Note that the arguments of both logarithms are exact exponents. 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 flip the second logarithm:

[Figure caption]

Since the product does not change from permutation of factors, we calmly multiplied four and two, and then figured out the 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 it down and get rid of the indicators:

[Figure caption]

Now let's get rid of the decimal logarithm by moving to a new base:

[Figure caption]

Basic logarithmic identity

Often in the process of solving it is required to represent a number as a logarithm to a given base. In this case, the formulas will help us:

In the first case, the number n becomes the exponent of the argument. Number n can be absolutely anything, because it's just the value of the logarithm.

The second formula is actually a paraphrased definition. It's called the basic logarithmic identity.

Indeed, what will happen if the number b raise to the power so that b to this extent gives a number a? That's right: this is the same number a. Read this paragraph carefully again - many people “hang” on it.

Like the new base conversion formulas, the basic logarithmic identity is sometimes the only possible solution.

Task. Find the value of the expression:

[Figure caption]

Note that log 25 64 = log 5 8 - just took out the square from the base and the argument of the logarithm. Given the rules for multiplying powers with the same base, we get:

[Figure caption]

If someone is not in the know, this was a real task from the exam :)

Logarithmic unit and logarithmic zero

In conclusion, I will give two identities that are difficult to call properties - rather, these are consequences from the definition of the logarithm. They are constantly found in problems and, surprisingly, create problems even for "advanced" students.

1. log a a= 1 is the logarithmic unit. Remember once and for all: the logarithm to any base a from this base itself is equal to one.
2. log a 1 = 0 is logarithmic zero. Base a can be anything, but if the argument is one, the logarithm is 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.

Let's start with properties of the logarithm of unity. 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 straightforward: since a 0 =1 for any a that satisfies the above conditions a>0 and a≠1 , then the proven equality log a 1=0 immediately follows from the definition of the logarithm.

Let's give examples of application of the considered property: log 3 1=0 , lg1=0 and .

Let's move on to the next property: the logarithm of a number equal to the base is equal to one, i.e, log a a=1 for a>0 , a≠1 . Indeed, since a 1 =a for any a , then by the definition of the logarithm log a a=1 .

Examples of using this property of logarithms are log 5 5=1 , log 5.6 5.6 and lne=1 .

For example, log 2 2 7 =7 , log10 -4 =-4 and .

Logarithm of the product of two positive numbers x and y is equal to the product of the 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 the 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 , whence the required equality follows by the definition of the logarithm.

Let's show examples of using the property of the logarithm of the product: log 5 (2 3)=log 5 2+log 5 3 and .

The product logarithm property 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 is easily proved.

For example, the natural logarithm of a 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 quotient logarithm property 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 proved like the formula for the logarithm of the product: since , then by the 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 degree. 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. We write this property of the logarithm of the degree in the form of a formula: log a b p =p log a |b|, where a>0 , a≠1 , b and p are numbers such that the degree of b p makes sense and b p >0 .

We first 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 power property, is equal to a p log a b . So we arrive at the equality b p =a p log a b , from which, by the definition of the 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|, whence 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 root of the nth degree is equal to the product of the fraction 1/n and the logarithm of the root 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 ), which is valid for any positive b , and the property of the logarithm of the degree: .

Here is an example of using this property: .

Now let's prove conversion formula to the new base of the logarithm kind . To do this, it suffices 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. Thus, the equality log c b=log a b log c a is proved, which means that the formula for the transition to a new base of the logarithm is also proved.

Let's show a couple of examples of applying 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 go to natural or decimal logarithms so that you can calculate the value of the logarithm from the table of logarithms. The formula for the transition to a new base of the logarithm also allows in some cases to find the value of a given logarithm, when the values ​​of some logarithms with other bases are known.

Often used is a special case of the formula for the transition to a new base of the logarithm for c=b of the form . This shows that log a b and log b a – . For example, .

Also often used is the formula , which is useful for finding logarithm values. To confirm our words, we will show how the value of the logarithm of the form is calculated using it. We have . To prove the formula it is enough to use the transition formula to the new base of the logarithm a: .

It remains to prove the comparison properties of logarithms.

Let us prove that for any positive numbers b 1 and b 2 , b 1 log a b 2 , and for a>1, the inequality log a b 1

Finally, it remains to prove the last of the listed properties of logarithms. We confine ourselves to proving its first part, that is, we prove that if a 1 >1 , a 2 >1 and a 1 1 is true log a 1 b>log a 2 b . The remaining statements of this property of logarithms are proved by a similar principle.

Let's use the opposite method. Suppose that for a 1 >1 , a 2 >1 and a 1 1 log a 1 b≤log a 2 b is true. By 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 the same bases, 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 . Thus, we have arrived at a contradiction to the condition a 1

Bibliography.

• Kolmogorov A.N., Abramov A.M., Dudnitsyn Yu.P. and others. Algebra and the Beginnings of Analysis: A Textbook for Grades 10-11 of General Educational Institutions.
• Gusev V.A., Mordkovich A.G. Mathematics (a manual for applicants to technical schools).