Instructions
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 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 the 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 an irrational equation and a rational one? 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 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 form of the integrand which of the table integrals fits in 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. By differentiating this expression, find the new differential in . So you will get new look 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 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 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.
Today we will learn how to solve the simplest logarithmic equations, where no preliminary transformations or selection of roots are required. But if you learn to solve such equations, then it will be much easier.
The simplest logarithmic equation is an equation of the form log a f (x) = b, where a, b are numbers (a > 0, a ≠ 1), f (x) is a certain function.
A distinctive feature of all logarithmic equations is the presence of the variable x under the logarithm sign. If this is the equation initially given in the problem, it is called the simplest. Any other logarithmic equations are reduced to the simplest by special transformations (see “Basic properties of logarithms”). However, numerous subtleties must be taken into account: extra roots may arise, so complex logarithmic equations will be considered separately.
How to solve such equations? It is enough to replace the number to the right of the equal sign with a logarithm in the same base as to the left. Then you can get rid of the sign of the logarithm. We get:
log a f (x) = b ⇒ log a f (x) = log a a b ⇒ f (x) = a b
We got the usual equation. Its roots are the roots of the original equation.
Taking out degrees
Often logarithmic equations, which outwardly look complex and threatening, are solved literally in a couple of lines without involving complex formulas. Today we will look at just such problems, where all that is required of you is to carefully reduce the formula to the canonical form and not get confused when searching for the domain of definition of logarithms.
Today, as you probably guessed from the title, we will solve logarithmic equations using formulas for the transition to the canonical form. The main “trick” of this video lesson will be working with degrees, or rather, deducing the degree from the basis and argument. Let's look at the rule:
Similarly, you can derive the degree from the base:
As we can see, if when we remove the degree from the argument of the logarithm we simply have an additional factor in front, then when we remove the degree from the base we get not just a factor, but an inverted factor. This needs to be remembered.
Finally, the most interesting thing. These formulas can be combined, then we get:
Of course, when making these transitions, there are certain pitfalls associated with the possible expansion of the scope of definition or, conversely, narrowing of the scope of definition. Judge for yourself:
log 3 x 2 = 2 ∙ log 3 x
If in the first case x could be any number other than 0, i.e. the requirement x ≠ 0, then in the second case we are satisfied with only x, which are not only not equal, but strictly greater than 0, because the domain of definition of the logarithm is that the argument be strictly greater than 0. Therefore, I will remind you of a wonderful formula from the 8th-9th grade algebra course:
That is, we must write our formula as follows:
log 3 x 2 = 2 ∙ log 3 |x |
Then no narrowing of the scope of definition will occur.
However, in today's video tutorial there will be no squares. If you look at our tasks, you will only see the roots. Therefore, we will not apply this rule, but you still need to keep it in mind so that at the right moment, when you see quadratic function in an argument or base of a logarithm, you will remember this rule and perform all transformations correctly.
So the first equation is:
To solve this problem, I propose to carefully look at each of the terms present in the formula.
Let's rewrite the first term as a power with a rational exponent:
We look at the second term: log 3 (1 − x). There is no need to do anything here, everything is already transformed here.
Finally, 0, 5. As I said in previous lessons, when solving logarithmic equations and formulas, I highly recommend moving from decimal fractions to ordinary ones. Let's do this:
0,5 = 5/10 = 1/2
Let's rewrite our original formula taking into account the resulting terms:
log 3 (1 − x ) = 1
Now let's move on to the canonical form:
log 3 (1 − x ) = log 3 3
We get rid of the logarithm sign by equating the arguments:
1 − x = 3
−x = 2
x = −2
That's it, we've solved the equation. However, let's still play it safe and find the domain of definition. To do this, let's go back to the original formula and see:
1 − x > 0
−x > −1
x< 1
Our root x = −2 satisfies this requirement, therefore x = −2 is a solution to the original equation. Now we have received a strict, clear justification. That's it, problem solved.
Let's move on to the second task:
Let's look at each term separately.
Let's write out the first one:
We have transformed the first term. We work with the second term:
Finally, the last term, which is to the right of the equal sign:
We substitute the resulting expressions instead of the terms in the resulting formula:
log 3 x = 1
Let's move on to the canonical form:
log 3 x = log 3 3
We get rid of the logarithm sign, equating the arguments, and we get:
x = 3
Again, just to be on the safe side, let's go back to the original equation and take a look. In the original formula, the variable x is present only in the argument, therefore,
x > 0
In the second logarithm, x is under the root, but again in the argument, therefore, the root must be greater than 0, i.e., the radical expression must be greater than 0. We look at our root x = 3. Obviously, it satisfies this requirement. Therefore, x = 3 is a solution to the original logarithmic equation. That's it, problem solved.
There are two key points in today's video tutorial:
1) do not be afraid to transform logarithms and, in particular, do not be afraid to take powers out of the sign of the logarithm, while remembering our basic formula: when removing a power from an argument, it is simply taken out without changes as a multiplier, and when removing a power from the base, this power is inverted.
2) the second point is related to the canonical form itself. We made the transition to the canonical form at the very end of the transformation of the logarithmic equation formula. Let me remind you of the following formula:
a = log b b a
Of course, by the expression “any number b”, I mean those numbers that satisfy the requirements imposed on the base of the logarithm, i.e.
1 ≠ b > 0
For such b, and since we already know the basis, this requirement will be fulfilled automatically. But for such b - any that satisfy this requirement — this transition can be done and we will succeed canonical form, in which you can get rid of the logarithm sign.
Expanding the domain of definition and extra roots
In the process of transforming logarithmic equations, an implicit expansion of the domain of definition may occur. Often students don’t even notice this, which leads to mistakes and incorrect answers.
Let's start with the simplest designs. The simplest logarithmic equation is the following:
log a f (x) = b
Note that x is present in only one argument of one logarithm. How do we solve such equations? We use the canonical form. To do this, imagine the number b = log a a b, and our equation will be rewritten as follows:
log a f (x) = log a a b
This entry is called the canonical form. It is to this that you should reduce any logarithmic equation that you will encounter not only in today’s lesson, but also in any independent and test work.
How to arrive at the canonical form and what techniques to use is a matter of practice. The main thing to understand is that as soon as you receive such a record, you can consider the problem solved. Because the next step is to write:
f (x) = a b
In other words, we get rid of the logarithm sign and simply equate the arguments.
Why all this talk? The fact is that the canonical form is applicable not only to the simplest problems, but also to any others. In particular, those that we will decide today. Let's see.
First task:
What is the problem with this equation? The fact is that the function is in two logarithms at once. The problem can be reduced to its simplest by simply subtracting one logarithm from another. But problems arise with the definition area: extra roots may appear. So let's just move one of the logarithms to the right:
This entry is much more similar to the canonical form. But there is one more nuance: in the canonical form, the arguments must be the same. And on the left we have the logarithm in base 3, and on the right in base 1/3. He knows that these bases need to be brought to the same number. For example, let's remember what negative powers are:
And then we’ll use the “−1” exponent outside of log as a multiplier:
Please note: the degree that was at the base is turned over and turns into a fraction. We got an almost canonical notation by getting rid of different bases, but in return we got the factor “−1” on the right. Let's factor this factor into the argument by turning it into a power:
Of course, having received the canonical form, we boldly cross out the sign of the logarithm and equate the arguments. At the same time, let me remind you that when raised to the power “−1”, the fraction is simply turned over - a proportion is obtained.
Let's use the basic property of proportion and multiply it crosswise:
(x − 4) (2x − 1) = (x − 5) (3x − 4)
2x 2 − x − 8x + 4 = 3x 2 − 4x − 15x + 20
2x 2 − 9x + 4 = 3x 2 − 19x + 20
x 2 − 10x + 16 = 0
We have before us the above quadratic equation, so we solve it using Vieta’s formulas:
(x − 8)(x − 2) = 0
x 1 = 8; x 2 = 2
That's all. Do you think the equation is solved? No! For such a solution we will receive 0 points, because in the original equation there are two logarithms with the variable x. Therefore, it is necessary to take into account the domain of definition.
And this is where the fun begins. Most students are confused: what is the domain of definition of a logarithm? Of course, all arguments (we have two) must be greater than zero:
(x − 4)/(3x − 4) > 0
(x − 5)/(2x − 1) > 0
Each of these inequalities must be solved, marked on a straight line, intersected, and only then seen which roots lie at the intersection.
I’ll be honest: this technique has a right to exist, it is reliable, and you will get the correct answer, but there are too many unnecessary steps in it. So let's go through our solution again and see: where exactly do we need to apply the scope? In other words, you need to clearly understand when exactly extra roots appear.
- Initially we had two logarithms. Then we moved one of them to the right, but this did not affect the definition area.
- Then we remove the power from the base, but there are still two logarithms, and in each of them there is a variable x.
- Finally, we cross out the signs of log and get the classic fractional rational equation.
It is at the last step that the scope of definition is expanded! As soon as we moved to a fractional-rational equation, getting rid of the log signs, the requirements for the variable x changed dramatically!
Consequently, the domain of definition can be considered not at the very beginning of the solution, but only at the mentioned step - before directly equating the arguments.
This is where the opportunity for optimization lies. On the one hand, we are required that both arguments be greater than zero. On the other hand, we further equate these arguments. Therefore, if at least one of them is positive, then the second one will also be positive!
So it turns out that requiring two inequalities to be fulfilled at once is overkill. It is enough to consider only one of these fractions. Which one exactly? The one that is simpler. For example, let's look at the right-hand fraction:
(x − 5)/(2x − 1) > 0
This is a typical fractional rational inequality; we solve it using the interval method:
How to place signs? Let's take a number that is obviously greater than all our roots. For example, 1 billion. And we substitute its fraction. We get a positive number, i.e. to the right of the root x = 5 there will be a plus sign.
Then the signs alternate, because there are no roots of even multiplicity anywhere. We are interested in intervals where the function is positive. Therefore, x ∈ (−∞; −1/2)∪(5; +∞).
Now let’s remember the answers: x = 8 and x = 2. Strictly speaking, these are not answers yet, but only candidates for the answer. Which one belongs to the specified set? Of course, x = 8. But x = 2 does not suit us in terms of its domain of definition.
In total, the answer to the first logarithmic equation will be x = 8. Now we have a competent, well-founded solution, taking into account the domain of definition.
Let's move on to the second equation:
log 5 (x − 9) = log 0.5 4 − log 5 (x − 5) + 3
Let me remind you that if there is a decimal fraction in the equation, then you should get rid of it. In other words, let's rewrite 0.5 as a common fraction. We immediately notice that the logarithm containing this base is easily calculated:
This is a very important moment! When we have degrees in both the base and the argument, we can derive the indicators of these degrees using the formula:
Let's go back to our original logarithmic equation and rewrite it:
log 5 (x − 9) = 1 − log 5 (x − 5)
We obtained a design quite close to the canonical form. However, we are confused by the terms and the minus sign to the right of the equal sign. Let's represent one as a logarithm to base 5:
log 5 (x − 9) = log 5 5 1 − log 5 (x − 5)
Subtract the logarithms on the right (in this case their arguments are divided):
log 5 (x − 9) = log 5 5/(x − 5)
Wonderful. So we got the canonical form! We cross out the log signs and equate the arguments:
(x − 9)/1 = 5/(x − 5)
This is a proportion that can easily be solved by multiplying crosswise:
(x − 9)(x − 5) = 5 1
x 2 − 9x − 5x + 45 = 5
x 2 − 14x + 40 = 0
Obviously, we have a reduced quadratic equation. It can be easily solved using Vieta's formulas:
(x − 10)(x − 4) = 0
x 1 = 10
x 2 = 4
We got two roots. But these are not final answers, but only candidates, because the logarithmic equation also requires checking the domain of definition.
I remind you: there is no need to search when every of the arguments will be greater than zero. It is enough to require that one argument—either x − 9 or 5/(x − 5)—be greater than zero. Consider the first argument:
x − 9 > 0
x > 9
Obviously, only x = 10 satisfies this requirement. This is the final answer. The whole problem is solved.
Once again, the key thoughts of today's lesson:
- As soon as the variable x appears in several logarithms, the equation ceases to be elementary, and its domain of definition must be calculated. Otherwise, you can easily write extra roots in the answer.
- Working with the domain itself can be significantly simplified if we write out the inequality not immediately, but exactly at the moment when we get rid of the log signs. After all, when the arguments are equated to each other, it is enough to require that only one of them be greater than zero.
Of course, we ourselves choose which argument to use to form an inequality, so it is logical to choose the simplest one. For example, in the second equation we chose the argument (x − 9), a linear function, as opposed to the fractional rational second argument. Agree, solving the inequality x − 9 > 0 is much easier than 5/(x − 5) > 0. Although the result is the same.
This remark greatly simplifies the search for ODZ, but be careful: you can use one inequality instead of two only if the arguments are precisely are equal to each other!
Of course, someone will now ask: what happens differently? Yes, it happens. For example, in the step itself, when we multiply two arguments containing a variable, there is a danger of unnecessary roots appearing.
Judge for yourself: first it is required that each of the arguments be greater than zero, but after multiplication it is enough that their product be greater than zero. As a result, the case where each of these fractions is negative is missed.
Therefore, if you are just starting to understand complex logarithmic equations, do not under any circumstances multiply logarithms containing the variable x - this will too often lead to the appearance of extra roots. It’s better to take one extra step, move one term to the other side and create a canonical form.
Well, what to do if you cannot do without multiplying such logarithms, we will discuss in the next video lesson. :)
Once again about the powers in the equation
Today we will examine a rather slippery topic concerning logarithmic equations, or more precisely, the removal of powers from the arguments and bases of logarithms.
I would even say that we will talk about the removal of even powers, because it is with even powers that most of the difficulties arise when solving real logarithmic equations.
Let's start with the canonical form. Let's say we have an equation of the form log a f (x) = b. In this case, we rewrite the number b using the formula b = log a a b . It turns out the following:
log a f (x) = log a a b
Then we equate the arguments:
f (x) = a b
The penultimate formula is called the canonical form. It is to this that they try to reduce any logarithmic equation, no matter how complex and scary it may seem at first glance.
So let's try it. Let's start with the first task:
Preliminary note: as I said, everything decimals in a logarithmic equation it is better to convert it into ordinary ones:
0,5 = 5/10 = 1/2
Let's rewrite our equation taking this fact into account. Note that both 1/1000 and 100 are powers of ten, and then let's take out powers wherever they are: from arguments and even from the base of logarithms:
And here many students have a question: “Where did the module on the right come from?” Indeed, why not simply write (x − 1)? Of course, now we will write (x − 1), but taking into account the domain of definition gives us the right to such a notation. After all, another logarithm already contains (x − 1), and this expression must be greater than zero.
But when we remove the square from the base of the logarithm, we must leave exactly the module at the base. Let me explain why.
The fact is that, from a mathematical point of view, taking a degree is tantamount to taking the root. In particular, when we square the expression (x − 1) 2, we are essentially taking the second root. But the square root is nothing more than a modulus. Exactly module, because even if the expression x − 1 is negative, when squared, the “minus” will still burn out. Further extraction of the root will give us a positive number - without any minuses.
In general, in order to avoid making offensive mistakes, remember once and for all:
The root of an even power of any function that is raised to the same power is equal not to the function itself, but to its modulus:
Let's return to our logarithmic equation. Speaking about the module, I argued that we can remove it painlessly. This is true. Now I will explain why. Strictly speaking, we had to consider two options:
- x − 1 > 0 ⇒ |x − 1| = x − 1
- x − 1< 0 ⇒ |х − 1| = −х + 1
Each of these options would need to be addressed. But there is one catch: the original formula already contains the function (x − 1) without any modulus. And following the domain of definition of logarithms, we have the right to immediately write that x − 1 > 0.
This requirement must be satisfied regardless of any modules and other transformations that we perform in the solution process. Therefore, there is no point in considering the second option - it will never arise. Even if we get some numbers when solving this branch of inequality, they still will not be included in the final answer.
Now we are literally one step away from the canonical form of the logarithmic equation. Let's represent the unit as follows:
1 = log x − 1 (x − 1) 1
In addition, we introduce the factor −4, which is on the right, into the argument:
log x − 1 10 −4 = log x − 1 (x − 1)
Before us is the canonical form of the logarithmic equation. We get rid of the logarithm sign:
10 −4 = x − 1
But since the base was a function (and not a prime number), we additionally require that this function be greater than zero and not equal to one. The resulting system will be:
Since the requirement x − 1 > 0 is satisfied automatically (after all, x − 1 = 10 −4), one of the inequalities can be deleted from our system. The second condition can also be crossed out, because x − 1 = 0.0001< 1. Итого получаем:
x = 1 + 0.0001 = 1.0001
This is the only root that automatically satisfies all the requirements of the domain of definition of the logarithm (however, all requirements were eliminated as obviously fulfilled in the conditions of our problem).
So the second equation is:
3 log 3 x x = 2 log 9 x x 2
How is this equation fundamentally different from the previous one? If only by the fact that the bases of logarithms - 3x and 9x - are not natural powers of each other. Therefore, the transition we used in the previous solution is not possible.
Let's at least get rid of degrees. In our case, the only degree is in the second argument:
3 log 3 x x = 2 ∙ 2 log 9 x |x |
However, the modulus sign can be removed, because the variable x is also at the base, i.e. x > 0 ⇒ |x| = x. Let's rewrite our logarithmic equation:
3 log 3 x x = 4 log 9 x x
We have obtained logarithms in which the arguments are the same, but the bases are different. What to do next? There are many options here, but we will consider only two of them, which are the most logical, and most importantly, these are quick and understandable techniques for most students.
We have already considered the first option: in any unclear situation, convert logarithms with a variable base to some constant base. For example, to a deuce. The transition formula is simple:
Of course, the role of variable c should be a normal number: 1 ≠ c > 0. Let in our case c = 2. Now we have before us an ordinary fractional rational equation. We collect all the elements on the left:
Obviously, it is better to remove the log 2 x factor, since it is present in both the first and second fractions.
log 2 x = 0;
3 log 2 9x = 4 log 2 3x
We break each log into two terms:
log 2 9x = log 2 9 + log 2 x = 2 log 2 3 + log 2 x;
log 2 3x = log 2 3 + log 2 x
Let's rewrite both sides of the equality taking into account these facts:
3 (2 log 2 3 + log 2 x ) = 4 (log 2 3 + log 2 x )
6 log 2 3 + 3 log 2 x = 4 log 2 3 + 4 log 2 x
2 log 2 3 = log 2 x
Now all that remains is to enter a two under the sign of the logarithm (it will turn into a power: 3 2 = 9):
log 2 9 = log 2 x
Before us is the classic canonical form, we get rid of the logarithm sign and get:
As expected, this root turned out to be greater than zero. It remains to check the domain of definition. Let's look at the reasons:
But the root x = 9 satisfies these requirements. Therefore, it is the final decision.
The conclusion from this solution is simple: do not be afraid of long calculations! It’s just that at the very beginning we chose a new base at random - and this significantly complicated the process.
But then the question arises: what basis is optimal? I will talk about this in the second method.
Let's go back to our original equation:
3 log 3x x = 2 log 9x x 2
3 log 3x x = 2 ∙ 2 log 9x |x |
x > 0 ⇒ |x| = x
3 log 3 x x = 4 log 9 x x
Now let's think a little: what number or function would be the optimal basis? It's obvious that the best option there will be c = x - what is already in the arguments. In this case, the formula log a b = log c b /log c a will take the form:
In other words, the expression is simply reversed. In this case, the argument and the basis change places.
This formula is very useful and is very often used in solving complex logarithmic equations. However, there is one very serious pitfall when using this formula. If we substitute the variable x instead of the base, then restrictions are imposed on it that were not previously observed:
There was no such limitation in the original equation. Therefore, we should separately check the case when x = 1. Substitute this value into our equation:
3 log 3 1 = 4 log 9 1
We get the correct numerical equality. Therefore x = 1 is a root. We found exactly the same root in the previous method at the very beginning of the solution.
But now that we have separately considered this particular case, we safely assume that x ≠ 1. Then our logarithmic equation will be rewritten in the following form:
3 log x 9x = 4 log x 3x
We expand both logarithms using the same formula as before. Note that log x x = 1:
3 (log x 9 + log x x ) = 4 (log x 3 + log x x )
3 log x 9 + 3 = 4 log x 3 + 4
3 log x 3 2 − 4 log x 3 = 4 − 3
2 log x 3 = 1
So we came to the canonical form:
log x 9 = log x x 1
x=9
We got the second root. It satisfies the requirement x ≠ 1. Therefore, x = 9 along with x = 1 is the final answer.
As you can see, the volume of calculations has decreased slightly. But when solving a real logarithmic equation, the number of steps will be much less also because you are not required to describe each step in such detail.
The key rule of today's lesson is the following: if the problem contains an even degree, from which the root of the same degree is extracted, then the output will be a modulus. However, this module can be removed if you pay attention to the domain of definition of logarithms.
But be careful: after this lesson, most students think that they understand everything. But when solving real problems, they cannot reproduce the entire logical chain. As a result, the equation acquires unnecessary roots, and the answer turns out to be incorrect.
Algebra 11th grade
Topic: “Methods for solving logarithmic equations”
Lesson objectives:
educational: formation of knowledge about in different ways solving logarithmic equations, the ability to apply them in each specific situation and choose any method for solving;
developing: developing the skills to observe, compare, apply knowledge in a new situation, identify patterns, generalize; developing skills of mutual control and self-control;
educational: fostering a responsible attitude to educational work, attentive perception of the material in the lesson, and careful note-taking.
Lesson type: lesson on introducing new material.
“The invention of logarithms, while reducing the work of the astronomer, extended his life.”
French mathematician and astronomer P.S. Laplace
Lesson progress
I. Setting the lesson goal
The studied definition of logarithm, properties of logarithms and logarithmic function will allow us to solve logarithmic equations. All logarithmic equations, no matter how complex they are, are solved using uniform algorithms. We'll look at these algorithms in today's lesson. There are not many of them. If you master them, then any equation with logarithms will be feasible for each of you.
Write down the topic of the lesson in your notebook: “Methods for solving logarithmic equations.” I invite everyone to cooperate.
II. Updating of reference knowledge
Let's prepare to study the topic of the lesson. You solve each task and write down the answer; you don’t have to write the condition. Work in pairs.
1) For what values of x does the function make sense:
(Answers are checked for each slide and errors are sorted out)
2) Do the graphs of the functions coincide?
3) Rewrite the equalities as logarithmic equalities:
4) Write the numbers as logarithms with base 2:
5) Calculate:
6) Try to restore or supplement the missing elements in these equalities.
III. Introduction to new material
The following statement is displayed on the screen:
“The equation is the golden key that opens all mathematical sesames.”
Modern Polish mathematician S. Kowal
Try to formulate the definition of a logarithmic equation. (An equation containing an unknown under the logarithm sign).
Let's consider the simplest logarithmic equation:logAx = b(where a>0, a ≠ 1). Since the logarithmic function increases (or decreases) on the set of positive numbers and takes all real values, then by the root theorem it follows that for any b this equation has, and only one, solution, and a positive one.
Remember the definition of logarithm. (The logarithm of a number x to the base a is an indicator of the power to which the base a must be raised to obtain the number x). From the definition of logarithm it immediately follows that AV is such a solution.
Write down the title: Methods for solving logarithmic equations
1. By definition of logarithm.
This is how the simplest equations of the form are solved.
Let's consider No. 514(a)): Solve the equation
How do you propose to solve it? (By definition of logarithm)
Solution. , Hence 2x - 4 = 4; x = 4.
In this task, 2x - 4 > 0, since > 0, so no extraneous roots can appear, and there is no need to check. The condition 2x - 4 > 0 does not need to be written out in this task.
2. Potentization(transition from the logarithm of a given expression to this expression itself).
Let's consider No. 519(g): log5(x2+8)-log5(x+1)=3log5 2
What feature did you notice? (The bases are the same and the logarithms of the two expressions are equal.) What can be done? (Potentize).
It should be taken into account that any solution is contained among all x for which the logarithmic expressions are positive.
Solution: ODZ:
X2+8>0 is an unnecessary inequality
log5(x2+8) =log5 23+ log5(x+1)
log5(x2+8)= log5 (8 x+8)
Let's potentiate the original equation
we get the equation x2+8= 8x+8
Let's solve it: x2-8x=0
Answer: 0; 8
IN general view transition to an equivalent system:
Equation
(The system contains a redundant condition - one of the inequalities need not be considered).
Question for the class: Which of these three solutions did you like best? (Discussion of methods).
You have the right to decide in any way.
3. Introduction of a new variable.
Let's consider No. 520(g). .
What did you notice? (This is a quadratic equation with respect to log3x) Any suggestions? (Introduce a new variable)
Solution. ODZ: x > 0.
Let , then the equation takes the form:. Discriminant D > 0. Roots according to Vieta's theorem:.
Let's go back to the replacement: or.
Having solved the simplest logarithmic equations, we get:
Answer: 27;
4. Logarithm both sides of the equation.
Solve the equation:.
Solution: ODZ: x>0, take the logarithm of both sides of the equation in base 10:
Let's apply the property of the logarithm of a power:
(logx + 3) logx = 4
Let logx = y, then (y + 3)y = 4
, (D > 0) roots according to Vieta’s theorem: y1 = -4 and y2 = 1.
Let's return to the replacement, we get: lgx = -4,; lgx = 1, .
Answer: 0.0001; 10.
5. Reduction to one base.
No. 523(c). Solve the equation:
Solution: ODZ: x>0. Let's move on to base 3.
6. Functional-graphic method.
№ 509(d). Solve the equation graphically: = 3 - x.
How do you propose to solve? (Build graphs of two functions y = log2x and y = 3 - x using points and look for the abscissa of the points of intersection of the graphs).
Look at your solution on the slide.
There is a way to avoid making graphs . It is as follows : if one of the functions y = f(x) increases, and the other y = g(x) decreases on the interval X, then the equation f(x)= g(x) has at most one root on the interval X.
If there is a root, then it can be guessed.
In our case, the function increases for x>0, and the function y = 3 - x decreases for all values of x, including for x>0, which means that the equation has no more than one root. Note that at x = 2 the equation turns into a true equality, since .
“The correct application of methods can be learned
only by applying them to various examples».
Danish historian of mathematics G. G. Zeiten
IV. Homework
P. 39 consider example 3, solve No. 514(b), No. 529(b), No. 520(b), No. 523(b)
V. Summing up the lesson
What methods of solving logarithmic equations did we look at in class?
In the next lessons we will look at more complex equations. To solve them, the studied methods will be useful.
Last slide shown:
“What is more than anything in the world?
Space.
What is the wisest thing?
Time.
What's the best part?
Achieve what you want."
Thales
I wish everyone to achieve what they want. Thank you for your cooperation and understanding.
Everyone knows why mathematics is needed. However, many people need help deciding mathematical problems and equations. Before we tell you how to solve logarithmic equations, you need to understand what they are. Equations that contain an unknown at the base of the logarithm or under its sign are called logarithmic equations. Equations that have the form: logaX = b, or those that can be reduced to this form, are considered to be the simplest logarithmic equations.
The right decision
For the right decision such equations, it is necessary to remember the properties of any logarithmic function:
- set of real numbers (range)
- set of positive numbers (domain)
- in the case when “a” is greater than 1, the logarithmic function strictly increases; if it is less, the logarithmic function decreases
- with the given parameters: loga "a" equals 1, and loga 1 equals zero, you need to take into account that "a" will not be equal to 1, and will be greater than 0.
Correct solution of logarithmic equations directly depends on understanding the logarithm itself. Let's take an example: 5x=11. X is the number to which 5 must be raised to make 11. This number is called the logarithm of 11 to base 5 and is written as follows: x = log511. Thus, we were able to solve the exponential equation: 5x=11, obtaining the answer: x=log511.
Logarithmic equations
An equation that has logarithms is called a logarithmic equation. In this equation, the unknown variables, as well as expressions with them, are located inside the logarithms themselves. And nowhere else! Examples of logarithmic equations: log2x=16, log5(x3-7)=log5(3x), log3(x+3)+20=15log(x+5), etc. Do not forget that various expressions with x can only be within a given logarithm.
Getting rid of logarithms
Methods for solving logarithmic equations are applied in accordance with the problem at hand, and the solution process itself as a whole is a very difficult task. Let's start with elementary equations. The simplest logarithmic equations have the following form:
- logx-21=11
- log5 (70x-1)=2
- log5x=25
Solving a logarithmic equation involves moving from an equation with logarithms to an equation in which there are none. And in the simplest equations this can be done in one step. It is for this reason that they are called protozoa. For example, we need to solve the following equation: log5x = log52. For this we do not need special knowledge. IN in this example we need to get rid of logarithms, which spoil the whole picture for us. We remove logarithms and get: x=2. Thus, in the future it is necessary to remove unnecessary logarithms, if possible. After all, it is precisely this sequence that allows you to decide logarithmic inequalities and equations. In mathematics, such actions are usually called potentiation. But getting rid of logarithms in this way has its own rules. If logarithms do not have any coefficients (that is, they are specified by themselves), and also if they have the same numerical base, the logarithms can be removed.
Remember, after we eliminate the logarithms, we are left with a simplified equation. Let's solve another example:
log9 (5x-4)-log9x. We potentiate and we get:
- 5x-4=x
- 5x=x+4
As you can see, by removing the logarithms, we get the usual equation, which is no longer difficult to solve. Now you can move on to more complex examples: log9 (60x-1)=2. We need to refer to the logarithm (the number to which the base is raised, in our case 9) to obtain the sublogarithmic expression (60x-1). Our logarithm is equal to 2. Therefore: 92 = 60x-1. There is no more logarithm. We solve the resulting equation: 60x-1=59, x = 1.
We solved this example according to the meaning of the logarithm. It should be noted that from any given number you can make a logarithm of the required form. This method is very useful in solving inequalities and logarithmic equations. If you need to find the root in the equation, let's look at how this can be done: log5(18 – x) = log55
If in our equation both sides of the equation have logarithms that have the same base, then we can equate the expressions that appear under the signs of our logarithms. We remove the common basis: log5. We get a simple equation: 18 – x = 5, x = 13.
In fact, solving logarithmic equations is not that difficult. Even taking into account the fact that the properties of logarithmic equations can differ significantly, all the same, there are no unsolvable tasks. It is necessary to know the properties of the logarithm itself, as well as to be able to apply them correctly. There is no need to rush: we remember the above instructions and begin to solve the tasks. There is no need to be scared under any circumstances complex equation, You have all the necessary knowledge and resources to easily cope with any of them.
On this lesson We will repeat the basic theoretical facts about logarithms and consider solving the simplest logarithmic equations.
Let us recall the central definition - the definition of a logarithm. It is related to the decision exponential equation. This equation has a single root, it is called the logarithm of b to base a:
Definition:
The logarithm of b to base a is the exponent to which base a must be raised to get b.
Let us remind you basic logarithmic identity.
The expression (expression 1) is the root of the equation (expression 2). Substitute the value x from expression 1 instead of x into expression 2 and get the main logarithmic identity:
So we see that each value is associated with a value. We denote b by x(), c by y, and thus obtain a logarithmic function:
For example: 
Let us recall the basic properties of the logarithmic function.
Let us pay attention once again, here, since under the logarithm there can be a strictly positive expression, as the base of the logarithm.
Rice. 1. Graph of a logarithmic function in different bases
The graph of the function at is shown in black. Rice. 1. If the argument increases from zero to infinity, the function increases from minus to plus infinity.
The graph of the function at is shown in red. Rice. 1.
Properties of this function:
Scope: ;
Range of values: ;
The function is monotonic throughout its entire domain of definition. When monotonically (strictly) increases, higher value the argument corresponds to the larger value of the function. When monotonically (strictly) decreases, a larger value of the argument corresponds to a smaller value of the function.
The properties of the logarithmic function are the key to solving a variety of logarithmic equations.
Let's consider the simplest logarithmic equation; all other logarithmic equations, as a rule, are reduced to this form.
Since the bases of logarithms and the logarithms themselves are equal, the functions under the logarithm are also equal, but we must not miss the domain of definition. Only a positive number can appear under the logarithm, we have:
We found out that the functions f and g are equal, so it is enough to choose any one inequality to comply with the ODZ.
Thus, we have a mixed system in which there is an equation and an inequality:
As a rule, it is not necessary to solve an inequality; it is enough to solve the equation and substitute the found roots into the inequality, thus performing a check.
Let us formulate a method for solving the simplest logarithmic equations:
Equalize the bases of logarithms;
Equate sublogarithmic functions;
Perform check.
Let's look at specific examples.
Example 1 - solve the equation:
The bases of logarithms are initially equal, we have the right to equate sublogarithmic expressions, do not forget about the ODZ, we choose the first logarithm to compose the inequality:
Example 2 - solve the equation:
This equation differs from the previous one in that the bases of the logarithms are less than one, but this does not affect the solution in any way:
Let's find the root and substitute it into the inequality:
We received an incorrect inequality, which means that the found root does not satisfy the ODZ.
Example 3 - solve the equation:
The bases of logarithms are initially equal, we have the right to equate sublogarithmic expressions, do not forget about the ODZ, we choose the second logarithm to compose the inequality:
Let's find the root and substitute it into the inequality:
Obviously, only the first root satisfies the DD.