Solving most mathematical problems in one way or another involves transforming numerical, algebraic or functional expressions. The above applies especially to the decision. In the versions of the Unified State Exam in mathematics, this type of problem includes, in particular, task C3. Learning to solve C3 tasks is important not only for the purpose of successfully passing the Unified State Exam, but also for the reason that this skill will be useful when studying a mathematics course in high school.
When completing C3 tasks, you have to solve various types of equations and inequalities. Among them are rational, irrational, exponential, logarithmic, trigonometric, containing modules (absolute values), as well as combined ones. This article discusses the main types of exponential equations and inequalities, as well as various methods for solving them. Read about solving other types of equations and inequalities in the “” section in articles devoted to methods for solving C3 problems from the Unified State Examination in mathematics.
Before we begin to analyze specific exponential equations and inequalities, as a math tutor, I suggest you brush up on some theoretical material that we will need.
Exponential function
What is an exponential function?
Function of the form y = a x, Where a> 0 and a≠ 1 is called exponential function.
Basic properties of exponential function y = a x:
Graph of an Exponential Function
The graph of the exponential function is exponent:
Graphs of exponential functions (exponents)
Solving exponential equations
Indicative are called equations in which the unknown variable is found only in exponents of some powers.
For solutions exponential equations you need to know and be able to use the following simple theorem:
Theorem 1. Exponential equation a f(x) = a g(x) (Where a > 0, a≠ 1) is equivalent to the equation f(x) = g(x).
In addition, it is useful to remember the basic formulas and operations with degrees:
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Example 1. Solve the equation:
Solution: We use the above formulas and substitution:
The equation then becomes:
The discriminant of the resulting quadratic equation is positive:
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This means that this equation has two roots. We find them:
Moving on to reverse substitution, we get:
The second equation has no roots, since the exponential function is strictly positive throughout the entire domain of definition. Let's solve the second one:
Taking into account what was said in Theorem 1, we move on to the equivalent equation: x= 3. This will be the answer to the task.
Answer: x = 3.
Example 2. Solve the equation:
Solution: The equation has no restrictions on the range of permissible values, since the radical expression makes sense for any value x(exponential function y = 9 4 -x positive and not equal to zero).
We solve the equation by equivalent transformations using the rules of multiplication and division of powers:
The last transition was carried out in accordance with Theorem 1.
Answer:x= 6.
Example 3. Solve the equation:
Solution: both sides of the original equation can be divided by 0.2 x. This transition will be equivalent, since this expression is greater than zero for any value x(the exponential function is strictly positive in its domain of definition). Then the equation takes the form:
Answer: x = 0.
Example 4. Solve the equation:
Solution: we simplify the equation to an elementary one by means of equivalent transformations using the rules of division and multiplication of powers given at the beginning of the article:
Dividing both sides of the equation by 4 x, as in the previous example, is an equivalent transformation, since this expression is not equal to zero for any values x.
Answer: x = 0.
Example 5. Solve the equation:
Solution: function y = 3x, standing on the left side of the equation, is increasing. Function y = —x The -2/3 on the right side of the equation is decreasing. This means that if the graphs of these functions intersect, then at most one point. In this case, it is easy to guess that the graphs intersect at the point x= -1. There will be no other roots.
Answer: x = -1.
Example 6. Solve the equation:
Solution: we simplify the equation by means of equivalent transformations, keeping in mind everywhere that the exponential function is strictly greater than zero for any value x and using the rules for calculating the product and quotient of powers given at the beginning of the article:
Answer: x = 2.
Solving exponential inequalities
Indicative are called inequalities in which the unknown variable is contained only in exponents of some powers.
For solutions exponential inequalities knowledge of the following theorem is required:
Theorem 2. If a> 1, then the inequality a f(x) > a g(x) is equivalent to an inequality of the same meaning: f(x) > g(x). If 0< a < 1, то показательное неравенство a f(x) > a g(x) is equivalent to an inequality with the opposite meaning: f(x) < g(x).
Example 7. Solve the inequality:
Solution: Let's present the original inequality in the form:
Let's divide both sides of this inequality by 3 2 x, in this case (due to the positivity of the function y= 3 2x) the inequality sign will not change:
Let's use the substitution:
Then the inequality will take the form:
So, the solution to the inequality is the interval:
moving to the reverse substitution, we get:
Due to the positivity of the exponential function, the left inequality is satisfied automatically. Using the well-known property of the logarithm, we move on to the equivalent inequality:
Since the base of the degree is a number greater than one, equivalent (by Theorem 2) is the transition to the following inequality:
So, we finally get answer:
Example 8. Solve the inequality:
Solution: Using the properties of multiplication and division of powers, we rewrite the inequality in the form:
Let's introduce a new variable:
Taking this substitution into account, the inequality takes the form:
Multiplying the numerator and denominator of the fraction by 7, we obtain the following equivalent inequality:
So, the following values of the variable satisfy the inequality t:
Then, moving to the reverse substitution, we get:
Since the base of the degree here is greater than one, the transition to the inequality will be equivalent (by Theorem 2):
Finally we get answer:
Example 9. Solve the inequality:
Solution:
We divide both sides of the inequality by the expression:
It is always greater than zero (due to the positivity of the exponential function), so there is no need to change the inequality sign. We get:
t located in the interval:
Moving on to the reverse substitution, we find that the original inequality splits into two cases:
The first inequality has no solutions due to the positivity of the exponential function. Let's solve the second one:
Example 10. Solve the inequality:
Solution:
Parabola branches y = 2x+2-x 2 are directed downwards, therefore it is limited from above by the value that it reaches at its vertex:
Parabola branches y = x 2 -2x The +2 in the indicator are directed upward, which means it is limited from below by the value that it reaches at its vertex:
At the same time, the function also turns out to be bounded from below y = 3 x 2 -2x+2, which is on the right side of the equation. It reaches its smallest value at the same point as the parabola in the exponent, and this value is 3 1 = 3. So, the original inequality can only be true if the function on the left and the function on the right take on the value , equal to 3 (the intersection of the ranges of values of these functions is only this number). This condition is satisfied at a single point x = 1.
Answer: x= 1.
In order to learn to decide exponential equations and inequalities, it is necessary to constantly train in solving them. Various teaching aids, problem books in elementary mathematics, collections of competitive problems, mathematics classes at school, as well as individual lessons with a professional tutor can help you in this difficult task. I sincerely wish you success in your preparation and excellent results in the exam.
Sergey Valerievich
P.S. Dear guests! Please do not write requests to solve your equations in the comments. Unfortunately, I have absolutely no time for this. Such messages will be deleted. Please read the article. Perhaps in it you will find answers to questions that did not allow you to solve your task on your own.
Let's look at how to solve exponential inequalities involving powers with different bases. The solution to such inequalities is similar to the solution to the corresponding ones.
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We group degrees with the same bases. It is more convenient to separate them on opposite sides of the inequality:
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From each pair of powers we take out of brackets the common factor - the power with the smaller exponent. Taking the common factor out of brackets means dividing each term by this factor. When dividing degrees with the same bases, we leave the base the same and subtract the exponents:
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You can immediately divide by 20 (20=4∙5), but practice shows that dividing in two stages allows you to avoid possible errors:
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Since the base is 2/5<1, показательная функция
decreases, therefore the sign of inequality between exponents changes to the opposite:
Let's solve the quadratic inequality using the interval method. The zeros of the function on the left side of the inequality are x1=-1; x2=2. We mark them on the number line.
To check the sign, take a zero: 0²-0-2=-2, in the interval to which the zero belongs, put “-“. We arrange the remaining signs in a checkerboard pattern. Since we are solving an inequality in which the left side is less than zero, we choose the interval with the “-” sign.
Answer: x ∈ (-1; 2).
A variant of inequalities of this type is that all powers have the same bases, but differ in the coefficients of x in the exponents.
On the left side we put out of brackets the degree with the lowest exponent
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We arrived at an exponential inequality. Since the base 7>1, the function
increases, the sign of inequality between the indicators does not change:
To solve this inequality using the interval method, we move all terms to the left side and reduce the fractions to
In this lesson we will look at various exponential inequalities and learn how to solve them, based on the technique for solving the simplest exponential inequalities
1. Definition and properties of an exponential function
Let us recall the definition and basic properties of the exponential function. The solution of all exponential equations and inequalities is based on these properties.
Exponential function is a function of the form , where the base is the degree and Here x is the independent variable, argument; y is the dependent variable, function.
Rice. 1. Graph of exponential function
The graph shows increasing and decreasing exponents, illustrating the exponential function with a base greater than one and less than one but greater than zero, respectively.
Both curves pass through the point (0;1)
Properties of the Exponential Function:
Domain: ;
Range of values: ;
The function is monotonic, increases with, decreases with.
A monotonic function takes each of its values given a single argument value.
When , when the argument increases from minus to plus infinity, the function increases from zero inclusive to plus infinity, i.e., for given values of the argument we have a monotonically increasing function (). On the contrary, when the argument increases from minus to plus infinity, the function decreases from infinity to zero inclusive, i.e., for given values of the argument we have a monotonically decreasing function ().
2. The simplest exponential inequalities, solution method, example
Based on the above, we present a method for solving simple exponential inequalities:
Technique for solving inequalities:
Equalize the bases of degrees;
Compare indicators by maintaining or changing the inequality sign to the opposite one.
The solution to complex exponential inequalities usually consists in reducing them to the simplest exponential inequalities.
The base of the degree is greater than one, which means the inequality sign is preserved:
Let's transform the right-hand side according to the properties of the degree:
The base of the degree is less than one, the inequality sign must be reversed:
To solve the quadratic inequality, we solve the corresponding quadratic equation:
Using Vieta's theorem we find the roots:
The branches of the parabola are directed upward.
Thus, we have a solution to the inequality:
It’s easy to guess that the right side can be represented as a power with an exponent of zero:
The base of the degree is greater than one, the inequality sign does not change, we get:
Let us recall the technique for solving such inequalities.
Consider the fractional-rational function:
We find the domain of definition:
Finding the roots of the function:
The function has a single root, 
We select intervals of constant sign and determine the signs of the function on each interval:
Rice. 2. Intervals of constancy of sign
Thus, we received the answer.
Answer: 
3. Solving standard exponential inequalities
Let's consider inequalities with the same indicators, but different bases.
One of the properties of the exponential function is that it takes strictly positive values for any value of the argument, which means that it can be divided into an exponential function. Let us divide the given inequality by its right side:
The base of the degree is greater than one, the inequality sign is preserved.
Let's illustrate the solution:
Figure 6.3 shows graphs of functions and . Obviously, when the argument is greater than zero, the graph of the function is higher, this function is larger. When the argument values are negative, the function goes lower, it is smaller. If the argument is equal, the functions are equal, which means that this point is also a solution to the given inequality.
Rice. 3. Illustration for example 4
Let us transform the given inequality according to the properties of the degree:
Here are some similar terms:
Let's divide both parts into:
Now we continue to solve similarly to example 4, divide both parts by:
The base of the degree is greater than one, the inequality sign remains:
4. Graphical solution of exponential inequalities
Example 6 - Solve the inequality graphically:
Let's look at the functions on the left and right sides and build a graph for each of them.
The function is exponential and increases over its entire domain of definition, i.e., for all real values of the argument.
The function is linear and decreases over its entire domain of definition, i.e., for all real values of the argument.
If these functions intersect, that is, the system has a solution, then such a solution is unique and can be easily guessed. To do this, we iterate over integers ()
It is easy to see that the root of this system is:
Thus, the graphs of the functions intersect at a point with an argument equal to one.
Now we need to get an answer. The meaning of the given inequality is that the exponent must be greater than or equal to the linear function, that is, be higher or coincide with it. The answer is obvious: (Figure 6.4)
Rice. 4. Illustration for example 6
So, we looked at solving various standard exponential inequalities. Next we move on to consider more complex exponential inequalities.
Bibliography
Mordkovich A. G. Algebra and the beginnings of mathematical analysis. - M.: Mnemosyne. Muravin G. K., Muravin O. V. Algebra and the beginnings of mathematical analysis. - M.: Bustard. Kolmogorov A. N., Abramov A. M., Dudnitsyn Yu. P. et al. Algebra and the beginnings of mathematical analysis. - M.: Enlightenment.
Math. md. Mathematics-repetition. com. Diffur. kemsu. ru.
Homework
1. Algebra and the beginnings of analysis, grades 10-11 (A. N. Kolmogorov, A. M. Abramov, Yu. P. Dudnitsyn) 1990, No. 472, 473;
2. Solve the inequality:
3. Solve the inequality.
Many people think that exponential inequalities are something complex and incomprehensible. And that learning to solve them is almost a great art, which only the Chosen are able to comprehend...
Complete nonsense! Exponential inequalities are easy. And they are always resolved simply. Well, almost always. :)
Today we will look at this topic inside and out. This lesson will be very useful for those who are just beginning to understand this section of school mathematics. Let's start with simple problems and move on to more complex issues. There won’t be any hard work today, but what you read now will be enough to solve most of the inequalities in all kinds of tests and independent work. And on this exam of yours too.
As always, let's start with the definition. An exponential inequality is any inequality that contains an exponential function. In other words, it can always be reduced to an inequality of the form
\[((a)^(x)) \gt b\]
Where the role of $b$ can be an ordinary number, or maybe something tougher. Examples? Yes please:
\[\begin(align) & ((2)^(x)) \gt 4;\quad ((2)^(x-1))\le \frac(1)(\sqrt(2));\ quad ((2)^(((x)^(2))-7x+14)) \lt 16; \\ & ((0,1)^(1-x)) \lt 0.01;\quad ((2)^(\frac(x)(2))) \lt ((4)^(\frac (4)(x))). \\\end(align)\]
I think the meaning is clear: there is an exponential function $((a)^(x))$, it is compared with something, and then asked to find $x$. In particularly clinical cases, instead of the variable $x$, they can put some function $f\left(x \right)$ and thereby complicate the inequality a little. :)
Of course, in some cases the inequality may appear more severe. For example:
\[((9)^(x))+8 \gt ((3)^(x+2))\]
Or even this:
In general, the complexity of such inequalities can be very different, but in the end they still reduce to the simple construction $((a)^(x)) \gt b$. And we will somehow figure out such a construction (in especially clinical cases, when nothing comes to mind, logarithms will help us). Therefore, now we will teach you how to solve such simple constructions.
Solving simple exponential inequalities
Let's consider something very simple. For example, this:
\[((2)^(x)) \gt 4\]
Obviously, the number on the right can be rewritten as a power of two: $4=((2)^(2))$. Thus, the original inequality can be rewritten in a very convenient form:
\[((2)^(x)) \gt ((2)^(2))\]
And now my hands are itching to “cross out” the twos in the bases of powers in order to get the answer $x \gt 2$. But before crossing out anything, let’s remember the powers of two:
\[((2)^(1))=2;\quad ((2)^(2))=4;\quad ((2)^(3))=8;\quad ((2)^( 4))=16;...\]
As you can see, the larger the number in the exponent, the larger the output number. "Thanks, Cap!" - one of the students will exclaim. Is it any different? Unfortunately, it happens. For example:
\[((\left(\frac(1)(2) \right))^(1))=\frac(1)(2);\quad ((\left(\frac(1)(2) \ right))^(2))=\frac(1)(4);\quad ((\left(\frac(1)(2) \right))^(3))=\frac(1)(8 );...\]
Here, too, everything is logical: the greater the degree, the more times the number 0.5 is multiplied by itself (i.e., divided in half). Thus, the resulting sequence of numbers is decreasing, and the difference between the first and second sequence is only in the base:
- If the base of degree $a \gt 1$, then as the exponent $n$ increases, the number $((a)^(n))$ will also increase;
- And vice versa, if $0 \lt a \lt 1$, then as the exponent $n$ increases, the number $((a)^(n))$ will decrease.
Summarizing these facts, we obtain the most important statement on which the entire solution of exponential inequalities is based:
If $a \gt 1$, then the inequality $((a)^(x)) \gt ((a)^(n))$ is equivalent to the inequality $x \gt n$. If $0 \lt a \lt 1$, then the inequality $((a)^(x)) \gt ((a)^(n))$ is equivalent to the inequality $x \lt n$.
In other words, if the base is greater than one, you can simply remove it - the inequality sign will not change. And if the base is less than one, then it can also be removed, but at the same time you will have to change the inequality sign.
Please note that we have not considered the options $a=1$ and $a\le 0$. Because in these cases uncertainty arises. Let's say how to solve an inequality of the form $((1)^(x)) \gt 3$? One to any power will again give one - we will never get three or more. Those. there are no solutions.
With negative reasons everything is even more interesting. For example, consider this inequality:
\[((\left(-2 \right))^(x)) \gt 4\]
At first glance, everything is simple:
Right? But no! It is enough to substitute a couple of even and a couple of odd numbers instead of $x$ to make sure that the solution is incorrect. Take a look:
\[\begin(align) & x=4\Rightarrow ((\left(-2 \right))^(4))=16 \gt 4; \\ & x=5\Rightarrow ((\left(-2 \right))^(5))=-32 \lt 4; \\ & x=6\Rightarrow ((\left(-2 \right))^(6))=64 \gt 4; \\ & x=7\Rightarrow ((\left(-2 \right))^(7))=-128 \lt 4. \\\end(align)\]
As you can see, the signs alternate. But there are also fractional powers and other nonsense. How, for example, would you order to calculate $((\left(-2 \right))^(\sqrt(7)))$ (minus two to the power of seven)? No way!
Therefore, for definiteness, we assume that in all exponential inequalities (and equations, by the way, too) $1\ne a \gt 0$. And then everything is solved very simply:
\[((a)^(x)) \gt ((a)^(n))\Rightarrow \left[ \begin(align) & x \gt n\quad \left(a \gt 1 \right), \\ & x \lt n\quad \left(0 \lt a \lt 1 \right). \\\end(align) \right.\]
In general, remember the main rule once again: if the base in an exponential equation is greater than one, you can simply remove it; and if the base is less than one, it can also be removed, but the sign of inequality will change.
Examples of solutions
So, let's look at a few simple exponential inequalities:
\[\begin(align) & ((2)^(x-1))\le \frac(1)(\sqrt(2)); \\ & ((0,1)^(1-x)) \lt 0.01; \\ & ((2)^(((x)^(2))-7x+14)) \lt 16; \\ & ((0,2)^(1+((x)^(2))))\ge \frac(1)(25). \\\end(align)\]
The primary task in all cases is the same: to reduce the inequalities to the simplest form $((a)^(x)) \gt ((a)^(n))$. This is exactly what we will now do with each inequality, and at the same time we will repeat the properties of degrees and exponential functions. So, let's go!
\[((2)^(x-1))\le \frac(1)(\sqrt(2))\]
What can you do here? Well, on the left we already have an indicative expression - nothing needs to be changed. But on the right there is some kind of crap: a fraction, and even a root in the denominator!
However, let us remember the rules for working with fractions and powers:
\[\begin(align) & \frac(1)(((a)^(n)))=((a)^(-n)); \\ & \sqrt[k](a)=((a)^(\frac(1)(k))). \\\end(align)\]
What does it mean? First, we can easily get rid of the fraction by turning it into a power with a negative exponent. And secondly, since the denominator has a root, it would be nice to turn it into a power - this time with a fractional exponent.
Let's apply these actions sequentially to the right side of the inequality and see what happens:
\[\frac(1)(\sqrt(2))=((\left(\sqrt(2) \right))^(-1))=((\left(((2)^(\frac( 1)(3))) \right))^(-1))=((2)^(\frac(1)(3)\cdot \left(-1 \right)))=((2)^ (-\frac(1)(3)))\]
Don't forget that when raising a degree to a power, the exponents of these degrees add up. And in general, when working with exponential equations and inequalities, it is absolutely necessary to know at least the simplest rules for working with powers:
\[\begin(align) & ((a)^(x))\cdot ((a)^(y))=((a)^(x+y)); \\ & \frac(((a)^(x)))(((a)^(y)))=((a)^(x-y)); \\ & ((\left(((a)^(x)) \right))^(y))=((a)^(x\cdot y)). \\\end(align)\]
Actually, we just applied the last rule. Therefore, our original inequality will be rewritten as follows:
\[((2)^(x-1))\le \frac(1)(\sqrt(2))\Rightarrow ((2)^(x-1))\le ((2)^(-\ frac(1)(3)))\]
Now we get rid of the two at the base. Since 2 > 1, the inequality sign will remain the same:
\[\begin(align) & x-1\le -\frac(1)(3)\Rightarrow x\le 1-\frac(1)(3)=\frac(2)(3); \\ & x\in \left(-\infty ;\frac(2)(3) \right]. \\\end(align)\]
That's the solution! The main difficulty is not at all in the exponential function, but in the competent transformation of the original expression: you need to carefully and quickly bring it to its simplest form.
Consider the second inequality:
\[((0.1)^(1-x)) \lt 0.01\]
So-so. Decimal fractions await us here. As I have said many times, in any expressions with powers you should get rid of decimals - this is often the only way to see a quick and simple solution. Here we will get rid of:
\[\begin(align) & 0.1=\frac(1)(10);\quad 0.01=\frac(1)(100)=((\left(\frac(1)(10) \ right))^(2)); \\ & ((0,1)^(1-x)) \lt 0,01\Rightarrow ((\left(\frac(1)(10) \right))^(1-x)) \lt ( (\left(\frac(1)(10) \right))^(2)). \\\end(align)\]
Here again we have the simplest inequality, and even with a base of 1/10, i.e. less than one. Well, we remove the bases, simultaneously changing the sign from “less” to “more”, and we get:
\[\begin(align) & 1-x \gt 2; \\ & -x \gt 2-1; \\ & -x \gt 1; \\& x \lt -1. \\\end(align)\]
We received the final answer: $x\in \left(-\infty ;-1 \right)$. Please note: the answer is precisely a set, and in no case a construction of the form $x \lt -1$. Because formally, such a construction is not a set at all, but an inequality with respect to the variable $x$. Yes, it is very simple, but it is not the answer!
Important Note. This inequality could be solved in another way - by reducing both sides to a power with a base greater than one. Take a look:
\[\frac(1)(10)=((10)^(-1))\Rightarrow ((\left(((10)^(-1)) \right))^(1-x)) \ lt ((\left(((10)^(-1)) \right))^(2))\Rightarrow ((10)^(-1\cdot \left(1-x \right))) \lt ((10)^(-1\cdot 2))\]
After such a transformation, we will again obtain an exponential inequality, but with a base of 10 > 1. This means that we can simply cross out the ten - the sign of the inequality will not change. We get:
\[\begin(align) & -1\cdot \left(1-x \right) \lt -1\cdot 2; \\ & x-1 \lt -2; \\ & x \lt -2+1=-1; \\ & x \lt -1. \\\end(align)\]
As you can see, the answer was exactly the same. At the same time, we saved ourselves from the need to change the sign and generally remember any rules. :)
\[((2)^(((x)^(2))-7x+14)) \lt 16\]
However, don't let this scare you. No matter what is in the indicators, the technology for solving inequality itself remains the same. Therefore, let us first note that 16 = 2 4. Let's rewrite the original inequality taking this fact into account:
\[\begin(align) & ((2)^(((x)^(2))-7x+14)) \lt ((2)^(4)); \\ & ((x)^(2))-7x+14 \lt 4; \\ & ((x)^(2))-7x+10 \lt 0. \\\end(align)\]
Hooray! We got the usual quadratic inequality! The sign has not changed anywhere, since the base is two - a number greater than one.
Zeros of a function on the number line
We arrange the signs of the function $f\left(x \right)=((x)^(2))-7x+10$ - obviously, its graph will be a parabola with branches up, so there will be “pluses” on the sides. We are interested in the region where the function is less than zero, i.e. $x\in \left(2;5 \right)$ is the answer to the original problem.
Finally, consider another inequality:
\[((0,2)^(1+((x)^(2))))\ge \frac(1)(25)\]
Again we see an exponential function with a decimal fraction at the base. Let's convert this fraction to a common fraction:
\[\begin(align) & 0.2=\frac(2)(10)=\frac(1)(5)=((5)^(-1))\Rightarrow \\ & \Rightarrow ((0 ,2)^(1+((x)^(2))))=((\left(((5)^(-1)) \right))^(1+((x)^(2) )))=((5)^(-1\cdot \left(1+((x)^(2)) \right)))\end(align)\]
In this case, we used the remark given earlier - we reduced the base to the number 5 > 1 in order to simplify our further solution. Let's do the same with the right side:
\[\frac(1)(25)=((\left(\frac(1)(5) \right))^(2))=((\left(((5)^(-1)) \ right))^(2))=((5)^(-1\cdot 2))=((5)^(-2))\]
Let us rewrite the original inequality taking into account both transformations:
\[((0,2)^(1+((x)^(2))))\ge \frac(1)(25)\Rightarrow ((5)^(-1\cdot \left(1+ ((x)^(2)) \right)))\ge ((5)^(-2))\]
The bases on both sides are the same and exceed one. There are no other terms on the right and left, so we simply “cross out” the fives and get a very simple expression:
\[\begin(align) & -1\cdot \left(1+((x)^(2)) \right)\ge -2; \\ & -1-((x)^(2))\ge -2; \\ & -((x)^(2))\ge -2+1; \\ & -((x)^(2))\ge -1;\quad \left| \cdot \left(-1 \right) \right. \\ & ((x)^(2))\le 1. \\\end(align)\]
This is where you need to be more careful. Many students like to simply take the square root of both sides of the inequality and write something like $x\le 1\Rightarrow x\in \left(-\infty ;-1 \right]$. Under no circumstances should this be done, since the root of an exact square is a modulus, and in no case an original variable:
\[\sqrt(((x)^(2)))=\left| x\right|\]
However, working with modules is not the most pleasant experience, is it? So we won't work. Instead, we simply move all the terms to the left and solve the usual inequality using the interval method:
$\begin(align) & ((x)^(2))-1\le 0; \\ & \left(x-1 \right)\left(x+1 \right)\le 0 \\ & ((x)_(1))=1;\quad ((x)_(2)) =-1; \\\end(align)$
We again mark the obtained points on the number line and look at the signs:
Please note: the dots are shaded Since we were solving a non-strict inequality, all points on the graph are shaded. Therefore, the answer will be: $x\in \left[ -1;1 \right]$ is not an interval, but a segment.
In general, I would like to note that there is nothing complicated about exponential inequalities. The meaning of all the transformations that we performed today comes down to a simple algorithm:
- Find the basis to which we will reduce all degrees;
- Carefully perform the transformations to obtain an inequality of the form $((a)^(x)) \gt ((a)^(n))$. Of course, instead of the variables $x$ and $n$ there can be much more complex functions, but the meaning will not change;
- Cross out the bases of degrees. In this case, the inequality sign may change if the base $a \lt 1$.
In fact, this is a universal algorithm for solving all such inequalities. And everything else they will tell you on this topic is just specific techniques and tricks that will simplify and speed up the transformation. We’ll talk about one of these techniques now. :)
Rationalization method
Let's consider another set of inequalities:
\[\begin(align) & ((\text( )\!\!\pi\!\!\text( ))^(x+7)) \gt ((\text( )\!\!\pi \!\!\text( ))^(((x)^(2))-3x+2)); \\ & ((\left(2\sqrt(3)-3 \right))^(((x)^(2))-2x)) \lt 1; \\ & ((\left(\frac(1)(3) \right))^(((x)^(2))+2x)) \gt ((\left(\frac(1)(9) \right))^(16-x)); \\ & ((\left(3-2\sqrt(2) \right))^(3x-((x)^(2)))) \lt 1. \\\end(align)\]
So what's so special about them? They're light. Although, stop! Is the number π raised to some power? What nonsense?
How to raise the number $2\sqrt(3)-3$ to a power? Or $3-2\sqrt(2)$? The problem writers obviously drank too much Hawthorn before sitting down to work. :)
In fact, there is nothing scary about these tasks. Let me remind you: an exponential function is an expression of the form $((a)^(x))$, where the base $a$ is any positive number except one. The number π is positive - we already know that. The numbers $2\sqrt(3)-3$ and $3-2\sqrt(2)$ are also positive - this is easy to see if you compare them with zero.
It turns out that all these “frightening” inequalities are solved no different from the simple ones discussed above? And are they resolved in the same way? Yes, that's absolutely right. However, using their example, I would like to consider one technique that greatly saves time on independent work and exams. We will talk about the method of rationalization. So, attention:
Any exponential inequality of the form $((a)^(x)) \gt ((a)^(n))$ is equivalent to the inequality $\left(x-n \right)\cdot \left(a-1 \right) \gt 0 $.
That's the whole method. :) Did you think that there would be some kind of another game? Nothing like this! But this simple fact, written literally in one line, will greatly simplify our work. Take a look:
\[\begin(matrix) ((\text( )\!\!\pi\!\!\text( ))^(x+7)) \gt ((\text( )\!\!\pi\ !\!\text( ))^(((x)^(2))-3x+2)) \\ \Downarrow \\ \left(x+7-\left(((x)^(2)) -3x+2 \right) \right)\cdot \left(\text( )\!\!\pi\!\!\text( )-1 \right) \gt 0 \\\end(matrix)\]
So there are no more exponential functions! And you don’t have to remember whether the sign changes or not. But a new problem arises: what to do with the damn multiplier \[\left(\text( )\!\!\pi\!\!\text( )-1 \right)\]? We don’t know what the exact value of the number π is. However, the captain seems to hint at the obvious:
\[\text( )\!\!\pi\!\!\text( )\approx 3.14... \gt 3\Rightarrow \text( )\!\!\pi\!\!\text( )-1\gt 3-1=2\]
In general, the exact value of π does not really concern us - it is only important for us to understand that in any case $\text( )\!\!\pi\!\!\text( )-1 \gt 2$, t .e. this is a positive constant, and we can divide both sides of the inequality by it:
\[\begin(align) & \left(x+7-\left(((x)^(2))-3x+2 \right) \right)\cdot \left(\text( )\!\! \pi\!\!\text( )-1 \right) \gt 0 \\ & x+7-\left(((x)^(2))-3x+2 \right) \gt 0; \\ & x+7-((x)^(2))+3x-2 \gt 0; \\ & -((x)^(2))+4x+5 \gt 0;\quad \left| \cdot \left(-1 \right) \right. \\ & ((x)^(2))-4x-5 \lt 0; \\ & \left(x-5 \right)\left(x+1 \right) \lt 0. \\\end(align)\]
As you can see, at a certain moment we had to divide by minus one - and the sign of inequality changed. At the end, I expanded the quadratic trinomial using Vieta's theorem - it is obvious that the roots are equal to $((x)_(1))=5$ and $((x)_(2))=-1$. Then everything is solved using the classical interval method:
Solving inequality using the interval method All points are removed because the original inequality is strict. We are interested in the region with negative values, so the answer is $x\in \left(-1;5 \right)$. That's the solution. :)
Let's move on to the next task:
\[((\left(2\sqrt(3)-3 \right))^(((x)^(2))-2x)) \lt 1\]
Everything here is generally simple, because there is a unit on the right. And we remember that one is any number raised to the zero power. Even if this number is an irrational expression at the base on the left:
\[\begin(align) & ((\left(2\sqrt(3)-3 \right))^(((x)^(2))-2x)) \lt 1=((\left(2 \sqrt(3)-3 \right))^(0)); \\ & ((\left(2\sqrt(3)-3 \right))^(((x)^(2))-2x)) \lt ((\left(2\sqrt(3)-3 \right))^(0)); \\\end(align)\]
Well, let's rationalize:
\[\begin(align) & \left(((x)^(2))-2x-0 \right)\cdot \left(2\sqrt(3)-3-1 \right) \lt 0; \\ & \left(((x)^(2))-2x-0 \right)\cdot \left(2\sqrt(3)-4 \right) \lt 0; \\ & \left(((x)^(2))-2x-0 \right)\cdot 2\left(\sqrt(3)-2 \right) \lt 0. \\\end(align)\ ]
All that remains is to figure out the signs. The factor $2\left(\sqrt(3)-2 \right)$ does not contain the variable $x$ - it is just a constant, and we need to find out its sign. To do this, note the following:
\[\begin(matrix) \sqrt(3) \lt \sqrt(4)=2 \\ \Downarrow \\ 2\left(\sqrt(3)-2 \right) \lt 2\cdot \left(2 -2 \right)=0 \\\end(matrix)\]
It turns out that the second factor is not just a constant, but a negative constant! And when dividing by it, the sign of the original inequality changes to the opposite:
\[\begin(align) & \left(((x)^(2))-2x-0 \right)\cdot 2\left(\sqrt(3)-2 \right) \lt 0; \\ & ((x)^(2))-2x-0 \gt 0; \\ & x\left(x-2 \right) \gt 0. \\\end(align)\]
Now everything becomes completely obvious. The roots of the square trinomial on the right are: $((x)_(1))=0$ and $((x)_(2))=2$. We mark them on the number line and look at the signs of the function $f\left(x \right)=x\left(x-2 \right)$:
The case when we are interested in side intervals We are interested in the intervals marked with a plus sign. All that remains is to write down the answer:
Let's move on to the next example:
\[((\left(\frac(1)(3) \right))^(((x)^(2))+2x)) \gt ((\left(\frac(1)(9) \ right))^(16-x))\]
Well, everything is completely obvious here: the bases contain powers of the same number. Therefore, I will write everything briefly:
\[\begin(matrix) \frac(1)(3)=((3)^(-1));\quad \frac(1)(9)=\frac(1)(((3)^( 2)))=((3)^(-2)) \\ \Downarrow \\ ((\left(((3)^(-1)) \right))^(((x)^(2) )+2x)) \gt ((\left(((3)^(-2)) \right))^(16-x)) \\\end(matrix)\]
\[\begin(align) & ((3)^(-1\cdot \left(((x)^(2))+2x \right))) \gt ((3)^(-2\cdot \ left(16-x \right))); \\ & ((3)^(-((x)^(2))-2x)) \gt ((3)^(-32+2x)); \\ & \left(-((x)^(2))-2x-\left(-32+2x \right) \right)\cdot \left(3-1 \right) \gt 0; \\ & -((x)^(2))-2x+32-2x \gt 0; \\ & -((x)^(2))-4x+32 \gt 0;\quad \left| \cdot \left(-1 \right) \right. \\ & ((x)^(2))+4x-32 \lt 0; \\ & \left(x+8 \right)\left(x-4 \right) \lt 0. \\\end(align)\]
As you can see, during the transformation process we had to multiply by a negative number, so the inequality sign changed. At the very end, I again applied Vieta's theorem to factor the quadratic trinomial. As a result, the answer will be the following: $x\in \left(-8;4 \right)$ - anyone can verify this by drawing a number line, marking the points and counting the signs. Meanwhile, we will move on to the last inequality from our “set”:
\[((\left(3-2\sqrt(2) \right))^(3x-((x)^(2)))) \lt 1\]
As you can see, at the base there is again an irrational number, and on the right there is again a unit. Therefore, we rewrite our exponential inequality as follows:
\[((\left(3-2\sqrt(2) \right))^(3x-((x)^(2)))) \lt ((\left(3-2\sqrt(2) \ right))^(0))\]
We apply rationalization:
\[\begin(align) & \left(3x-((x)^(2))-0 \right)\cdot \left(3-2\sqrt(2)-1 \right) \lt 0; \\ & \left(3x-((x)^(2))-0 \right)\cdot \left(2-2\sqrt(2) \right) \lt 0; \\ & \left(3x-((x)^(2))-0 \right)\cdot 2\left(1-\sqrt(2) \right) \lt 0. \\\end(align)\ ]
However, it is quite obvious that $1-\sqrt(2) \lt 0$, since $\sqrt(2)\approx 1,4... \gt 1$. Therefore, the second factor is again a negative constant, by which both sides of the inequality can be divided:
\[\begin(matrix) \left(3x-((x)^(2))-0 \right)\cdot 2\left(1-\sqrt(2) \right) \lt 0 \\ \Downarrow \ \\end(matrix)\]
\[\begin(align) & 3x-((x)^(2))-0 \gt 0; \\ & 3x-((x)^(2)) \gt 0;\quad \left| \cdot \left(-1 \right) \right. \\ & ((x)^(2))-3x \lt 0; \\ & x\left(x-3 \right) \lt 0. \\\end(align)\]
Move to another base
A separate problem when solving exponential inequalities is the search for the “correct” basis. Unfortunately, it is not always obvious at the first glance at a task what to take as a basis, and what to do according to the degree of this basis.
But don’t worry: there is no magic or “secret” technology here. In mathematics, any skill that cannot be algorithmized can be easily developed through practice. But for this you will have to solve problems of different levels of complexity. For example, like this:
\[\begin(align) & ((2)^(\frac(x)(2))) \lt ((4)^(\frac(4)(x))); \\ & ((\left(\frac(1)(3) \right))^(\frac(3)(x)))\ge ((3)^(2+x)); \\ & ((\left(0,16 \right))^(1+2x))\cdot ((\left(6,25 \right))^(x))\ge 1; \\ & ((\left(\frac(27)(\sqrt(3)) \right))^(-x)) \lt ((9)^(4-2x))\cdot 81. \\\ end(align)\]
Difficult? Scary? It's easier than hitting a chicken on the asphalt! Let's try. First inequality:
\[((2)^(\frac(x)(2))) \lt ((4)^(\frac(4)(x)))\]
Well, I think everything is clear here:
We rewrite the original inequality, reducing everything to base two:
\[((2)^(\frac(x)(2))) \lt ((2)^(\frac(8)(x)))\Rightarrow \left(\frac(x)(2)- \frac(8)(x) \right)\cdot \left(2-1 \right) \lt 0\]
Yes, yes, you heard it right: I just applied the rationalization method described above. Now we need to work carefully: we have a fractional-rational inequality (this is one that has a variable in the denominator), so before equating anything to zero, we need to bring everything to a common denominator and get rid of the constant factor.
\[\begin(align) & \left(\frac(x)(2)-\frac(8)(x) \right)\cdot \left(2-1 \right) \lt 0; \\ & \left(\frac(((x)^(2))-16)(2x) \right)\cdot 1 \lt 0; \\ & \frac(((x)^(2))-16)(2x) \lt 0. \\\end(align)\]
Now we use the standard interval method. Numerator zeros: $x=\pm 4$. The denominator goes to zero only when $x=0$. There are three points in total that need to be marked on the number line (all points are pinned out because the inequality sign is strict). We get:
More complex case: three roots As you might guess, the shading marks those intervals at which the expression on the left takes negative values. Therefore, the final answer will include two intervals at once:
The ends of the intervals are not included in the answer because the original inequality was strict. No further verification of this answer is required. In this regard, exponential inequalities are much simpler than logarithmic ones: no ODZ, no restrictions, etc.
Let's move on to the next task:
\[((\left(\frac(1)(3) \right))^(\frac(3)(x)))\ge ((3)^(2+x))\]
There are no problems here either, since we already know that $\frac(1)(3)=((3)^(-1))$, so the whole inequality can be rewritten as follows:
\[\begin(align) & ((\left(((3)^(-1)) \right))^(\frac(3)(x)))\ge ((3)^(2+x ))\Rightarrow ((3)^(-\frac(3)(x)))\ge ((3)^(2+x)); \\ & \left(-\frac(3)(x)-\left(2+x \right) \right)\cdot \left(3-1 \right)\ge 0; \\ & \left(-\frac(3)(x)-2-x \right)\cdot 2\ge 0;\quad \left| :\left(-2 \right) \right. \\ & \frac(3)(x)+2+x\le 0; \\ & \frac(((x)^(2))+2x+3)(x)\le 0. \\\end(align)\]
Please note: in the third line I decided not to waste time on trifles and immediately divide everything by (−2). Minul went into the first bracket (now there are pluses everywhere), and two was reduced with a constant factor. This is exactly what you should do when preparing real calculations for independent and test work - you don’t need to describe every action and transformation directly.
Next, the familiar method of intervals comes into play. Numerator zeros: but there are none. Because the discriminant will be negative. In turn, the denominator is reset only when $x=0$ - just like last time. Well, it is clear that to the right of $x=0$ the fraction will take positive values, and to the left - negative. Since we are interested in negative values, the final answer is: $x\in \left(-\infty ;0 \right)$.
\[((\left(0.16 \right))^(1+2x))\cdot ((\left(6.25 \right))^(x))\ge 1\]
What should you do with decimal fractions in exponential inequalities? That's right: get rid of them, converting them into ordinary ones. Here we will translate:
\[\begin(align) & 0.16=\frac(16)(100)=\frac(4)(25)\Rightarrow ((\left(0.16 \right))^(1+2x)) =((\left(\frac(4)(25) \right))^(1+2x)); \\ & 6.25=\frac(625)(100)=\frac(25)(4)\Rightarrow ((\left(6.25 \right))^(x))=((\left(\ frac(25)(4)\right))^(x)). \\\end(align)\]
So what did we get in the foundations of exponential functions? And we got two mutually inverse numbers:
\[\frac(25)(4)=((\left(\frac(4)(25) \right))^(-1))\Rightarrow ((\left(\frac(25)(4) \ right))^(x))=((\left(((\left(\frac(4)(25) \right))^(-1)) \right))^(x))=((\ left(\frac(4)(25) \right))^(-x))\]
Thus, the original inequality can be rewritten as follows:
\[\begin(align) & ((\left(\frac(4)(25) \right))^(1+2x))\cdot ((\left(\frac(4)(25) \right) )^(-x))\ge 1; \\ & ((\left(\frac(4)(25) \right))^(1+2x+\left(-x \right)))\ge ((\left(\frac(4)(25) \right))^(0)); \\ & ((\left(\frac(4)(25) \right))^(x+1))\ge ((\left(\frac(4)(25) \right))^(0) ). \\\end(align)\]
Of course, when multiplying powers with the same base, their exponents add up, which is what happened in the second line. In addition, we represented the unit on the right, also as a power in base 4/25. All that remains is to rationalize:
\[((\left(\frac(4)(25) \right))^(x+1))\ge ((\left(\frac(4)(25) \right))^(0)) \Rightarrow \left(x+1-0 \right)\cdot \left(\frac(4)(25)-1 \right)\ge 0\]
Note that $\frac(4)(25)-1=\frac(4-25)(25) \lt 0$, i.e. the second factor is a negative constant, and when dividing by it, the inequality sign will change:
\[\begin(align) & x+1-0\le 0\Rightarrow x\le -1; \\ & x\in \left(-\infty ;-1 \right]. \\\end(align)\]
Finally, the last inequality from the current “set”:
\[((\left(\frac(27)(\sqrt(3)) \right))^(-x)) \lt ((9)^(4-2x))\cdot 81\]
In principle, the idea of the solution here is also clear: all exponential functions included in the inequality must be reduced to base “3”. But for this you will have to tinker a little with roots and powers:
\[\begin(align) & \frac(27)(\sqrt(3))=\frac(((3)^(3)))(((3)^(\frac(1)(3)) ))=((3)^(3-\frac(1)(3)))=((3)^(\frac(8)(3))); \\ & 9=((3)^(2));\quad 81=((3)^(4)). \\\end(align)\]
Taking these facts into account, the original inequality can be rewritten as follows:
\[\begin(align) & ((\left(((3)^(\frac(8)(3))) \right))^(-x)) \lt ((\left(((3) ^(2))\right))^(4-2x))\cdot ((3)^(4)); \\ & ((3)^(-\frac(8x)(3))) \lt ((3)^(8-4x))\cdot ((3)^(4)); \\ & ((3)^(-\frac(8x)(3))) \lt ((3)^(8-4x+4)); \\ & ((3)^(-\frac(8x)(3))) \lt ((3)^(4-4x)). \\\end(align)\]
Pay attention to the 2nd and 3rd lines of the calculations: before doing anything with the inequality, be sure to bring it to the form that we talked about from the very beginning of the lesson: $((a)^(x)) \lt ((a)^(n))$. As long as you have some left-handed factors, additional constants, etc. on the left or right, no rationalization or “crossing out” of grounds can be performed! Countless tasks have been completed incorrectly due to a failure to understand this simple fact. I myself constantly observe this problem with my students when we are just starting to analyze exponential and logarithmic inequalities.
But let's return to our task. Let's try to do without rationalization this time. Let us remember: the base of the degree is greater than one, so the triples can simply be crossed out - the inequality sign will not change. We get:
\[\begin(align) & -\frac(8x)(3) \lt 4-4x; \\ & 4x-\frac(8x)(3) \lt 4; \\ & \frac(4x)(3) \lt 4; \\ & 4x \lt 12; \\ & x \lt 3. \\\end(align)\]
That's all. Final answer: $x\in \left(-\infty ;3 \right)$.
Isolating a stable expression and replacing a variable
In conclusion, I propose solving four more exponential inequalities, which are already quite difficult for unprepared students. To cope with them, you need to remember the rules for working with degrees. In particular, putting common factors out of brackets.
But the most important thing is to learn to understand what exactly can be taken out of brackets. Such an expression is called stable - it can be denoted by a new variable and thus get rid of the exponential function. So, let's look at the tasks:
\[\begin(align) & ((5)^(x+2))+((5)^(x+1))\ge 6; \\ & ((3)^(x))+((3)^(x+2))\ge 90; \\ & ((25)^(x+1.5))-((5)^(2x+2)) \gt 2500; \\ & ((\left(0.5 \right))^(-4x-8))-((16)^(x+1.5)) \gt 768. \\\end(align)\]
Let's start from the very first line. Let us write this inequality separately:
\[((5)^(x+2))+((5)^(x+1))\ge 6\]
Note that $((5)^(x+2))=((5)^(x+1+1))=((5)^(x+1))\cdot 5$, so the right-hand side can be rewrite:
Note that there are no other exponential functions except $((5)^(x+1))$ in the inequality. And in general, the variable $x$ does not appear anywhere else, so let’s introduce a new variable: $((5)^(x+1))=t$. We get the following construction:
\[\begin(align) & 5t+t\ge 6; \\&6t\ge 6; \\ & t\ge 1. \\\end(align)\]
We return to the original variable ($t=((5)^(x+1))$), and at the same time remember that 1=5 0 . We have:
\[\begin(align) & ((5)^(x+1))\ge ((5)^(0)); \\ & x+1\ge 0; \\ & x\ge -1. \\\end(align)\]
That's the solution! Answer: $x\in \left[ -1;+\infty \right)$. Let's move on to the second inequality:
\[((3)^(x))+((3)^(x+2))\ge 90\]
Everything is the same here. Note that $((3)^(x+2))=((3)^(x))\cdot ((3)^(2))=9\cdot ((3)^(x))$ . Then the left side can be rewritten:
\[\begin(align) & ((3)^(x))+9\cdot ((3)^(x))\ge 90;\quad \left| ((3)^(x))=t \right. \\&t+9t\ge 90; \\ & 10t\ge 90; \\ & t\ge 9\Rightarrow ((3)^(x))\ge 9\Rightarrow ((3)^(x))\ge ((3)^(2)); \\ & x\ge 2\Rightarrow x\in \left[ 2;+\infty \right). \\\end(align)\]
This is approximately how you need to draw up a solution for real tests and independent work.
Well, let's try something more complicated. For example, here is the inequality:
\[((25)^(x+1.5))-((5)^(2x+2)) \gt 2500\]
What's the problem here? First of all, the bases of the exponential functions on the left are different: 5 and 25. However, 25 = 5 2, so the first term can be transformed:
\[\begin(align) & ((25)^(x+1.5))=((\left(((5)^(2)) \right))^(x+1.5))= ((5)^(2x+3)); \\ & ((5)^(2x+3))=((5)^(2x+2+1))=((5)^(2x+2))\cdot 5. \\\end(align )\]
As you can see, at first we brought everything to the same base, and then we noticed that the first term can easily be reduced to the second - you just need to expand the exponent. Now you can safely introduce a new variable: $((5)^(2x+2))=t$, and the whole inequality will be rewritten as follows:
\[\begin(align) & 5t-t\ge 2500; \\&4t\ge 2500; \\ & t\ge 625=((5)^(4)); \\ & ((5)^(2x+2))\ge ((5)^(4)); \\ & 2x+2\ge 4; \\&2x\ge 2; \\ & x\ge 1. \\\end(align)\]
And again, no difficulties! Final answer: $x\in \left[ 1;+\infty \right)$. Let's move on to the final inequality in today's lesson:
\[((\left(0.5 \right))^(-4x-8))-((16)^(x+1.5)) \gt 768\]
The first thing you should pay attention to is, of course, the decimal fraction in the base of the first power. It is necessary to get rid of it, and at the same time bring all exponential functions to the same base - the number “2”:
\[\begin(align) & 0.5=\frac(1)(2)=((2)^(-1))\Rightarrow ((\left(0.5 \right))^(-4x- 8))=((\left(((2)^(-1)) \right))^(-4x-8))=((2)^(4x+8)); \\ & 16=((2)^(4))\Rightarrow ((16)^(x+1.5))=((\left(((2)^(4)) \right))^( x+1.5))=((2)^(4x+6)); \\ & ((2)^(4x+8))-((2)^(4x+6)) \gt 768. \\\end(align)\]
Great, we’ve taken the first step—everything has led to the same foundation. Now you need to select a stable expression. Note that $((2)^(4x+8))=((2)^(4x+6+2))=((2)^(4x+6))\cdot 4$. If we introduce a new variable $((2)^(4x+6))=t$, then the original inequality can be rewritten as follows:
\[\begin(align) & 4t-t \gt 768; \\ & 3t \gt 768; \\ & t \gt 256=((2)^(8)); \\ & ((2)^(4x+6)) \gt ((2)^(8)); \\ & 4x+6 \gt 8; \\ & 4x \gt 2; \\ & x \gt \frac(1)(2)=0.5. \\\end(align)\]
Naturally, the question may arise: how did we discover that 256 = 2 8? Unfortunately, here you just need to know the powers of two (and at the same time the powers of three and five). Well, or divide 256 by 2 (you can divide, since 256 is an even number) until we get the result. It will look something like this:
\[\begin(align) & 256=128\cdot 2= \\ & =64\cdot 2\cdot 2= \\ & =32\cdot 2\cdot 2\cdot 2= \\ & =16\cdot 2 \cdot 2\cdot 2\cdot 2= \\ & =8\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2= \\ & =4\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2= \\ & =2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2= \\ & =((2)^(8)).\end(align )\]
The same is true with three (the numbers 9, 27, 81 and 243 are its degrees), and with seven (the numbers 49 and 343 would also be nice to remember). Well, the five also has “beautiful” degrees that you need to know:
\[\begin(align) & ((5)^(2))=25; \\ & ((5)^(3))=125; \\ & ((5)^(4))=625; \\ & ((5)^(5))=3125. \\\end(align)\]
Of course, if you wish, all these numbers can be restored in your mind by simply multiplying them successively by each other. However, when you have to solve several exponential inequalities, and each next one is more difficult than the previous one, then the last thing you want to think about is the powers of some numbers. And in this sense, these problems are more complex than “classical” inequalities that are solved by the interval method.
I hope this lesson helped you in mastering this topic. If something is unclear, ask in the comments. And see you in the next lessons. :)
