exponential equations. More difficult cases. Solving exponential inequalities: basic methods

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 especially 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 look 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 come down to a simple construction $((a)^(x)) \gt b$. And we will somehow deal with such a design (in especially clinical cases, when nothing comes to mind, logarithms will help us). Therefore, now we will learn how to solve such simple constructions.

Solution of the simplest exponential inequalities

Let's look at something very simple. For example, here it is:

\[((2)^(x)) \gt 4\]

Obviously, the number on the right can be rewritten as a power of two: $4=((2)^(2))$. Thus, the original inequality is rewritten in a very convenient form:

\[((2)^(x)) \gt ((2)^(2))\]

And now the hands are itching to "cross out" the deuces, standing in the bases of the degrees, in order to get the answer $x \gt 2$. But before we cross 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. Does it happen differently? 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 (that is, it is divided in half). Thus, the resulting sequence of numbers decreases, and the difference between the first and second sequences is only in the base:

• If the base of degree $a \gt 1$, then as the exponent $n$ grows, the number $((a)^(n))$ will also grow;
• Conversely, if $0 \lt a \lt 1$, then as the exponent $n$ grows, the number $((a)^(n))$ will decrease.

Summing up these facts, we get 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 the sign of inequality will also have to be changed.

Note that we have not considered the $a=1$ and $a\le 0$ options. Because in these cases there is uncertainty. Suppose how to solve an inequality of the form $((1)^(x)) \gt 3$? A one to any power will again give a one - we will never get a three or more. Those. there are no solutions.

With negative bases, it's even more interesting. Consider, for example, the following 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 wrong. 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 still fractional degrees and other tin. How, for example, would you order to count $((\left(-2 \right))^(\sqrt(7)))$ (minus two raised to the root 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, once again remember the main rule: if the base in the 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 this will change the inequality sign.

Solution examples

So, consider 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 is the same in all cases: to reduce the inequalities to the simplest form $((a)^(x)) \gt ((a)^(n))$. This is what we will now do with each inequality, and at the same time we will repeat the properties of powers and the exponential function. So let's go!

\[((2)^(x-1))\le \frac(1)(\sqrt(2))\]

What can be done here? Well, on the left we already have a demonstrative 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, 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 negative exponent. And secondly, since the denominator is the root, it would be nice to turn it into a degree - this time with a fractional exponent.

Let's apply these actions sequentially to the right side of the inequality and see what happens:

\[\frac(1)(\sqrt(2))=((\left(\sqrt(2) \right))^(-1))=((\left(((2)^(\frac( 1)(3))) \right))^(-1))=((2)^(\frac(1)(3)\cdot \left(-1 \right)))=((2)^ (-\frac(1)(3)))\]

Do not forget that when raising a degree to a power, the exponents of these degrees are added. 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 deuce at the base. Since 2 > 1, the inequality sign remains 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 whole 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 as quickly as possible bring it to its simplest form.

Consider the second inequality:

\[((0,1)^(1-x)) \lt 0,01\]

So-so. Here we are waiting for decimal fractions. As I have said many times, in any expressions with powers, you should get rid of decimal fractions - often this is the only way to see a quick and easy solution. Here's what we'll get rid of:

\[\begin(align) & 0,1=\frac(1)(10);\quad 0,01=\frac(1)(100)=((\left(\frac(1)(10) \ right))^(2)); \\ & ((0,1)^(1-x)) \lt 0,01\Rightarrow ((\left(\frac(1)(10) \right))^(1-x)) \lt ( (\left(\frac(1)(10) \right))^(2)). \\\end(align)\]

Before us is again the simplest inequality, and even with the base 1/10, i.e. less than one. Well, we remove the bases, simultaneously changing the sign from "less" to "greater", and we get:

\[\begin(align) & 1-x \gt 2; \\ & -x \gt 2-1; \\ & -x \gt 1; \\& x \lt -1. \\\end(align)\]

We got the final answer: $x\in \left(-\infty ;-1 \right)$. Please note that the answer is exactly the set, and in no case is the 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's very simple, but it's not the answer!

Important note. This inequality could be solved in another way - by reducing both parts 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 again get an exponential inequality, but with a base of 10 > 1. And this means that you can simply cross out the ten - the inequality sign 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 is exactly the same. At the same time, we saved ourselves from the need to change the sign and generally remember some rules there. :)

\[((2)^(((x)^(2))-7x+14)) \lt 16\]

However, don't let that scare you. Whatever is in the indicators, the technology for solving the inequality itself remains the same. Therefore, we note first 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 square inequality! The sign has not changed anywhere, since the base is a deuce - a number greater than one.

Function zeros 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 in 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 took advantage of the remark made earlier - we reduced the base to the number 5\u003e 1 in order to simplify our further decision. 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's 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 greater than one. There are no other terms on the right and left, so we just “cross out” the fives and we 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 have to be careful. Many students like to simply take the square root of both sides of the inequality and write something like $x\le 1\Rightarrow x\in \left(-\infty ;-1 \right]$. You should never do this, since the root of the exact square is the modulus, and by no means the original variable:

\[\sqrt(((x)^(2)))=\left| x\right|\]

However, working with modules is not the most pleasant experience, right? So we won't work. Instead, we simply move all the terms to the left and solve the usual inequality using the interval method:

$\begin(align) & ((x)^(2))-1\le 0; \\ & \left(x-1 \right)\left(x+1 \right)\le 0 \\ & ((x)_(1))=1;\quad ((x)_(2)) =-1; \\\end(align)$

Again, we mark the obtained points on the number line and look at the signs:

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 in exponential inequalities. The meaning of all the transformations that we performed today boils down to a simple algorithm:

• Find the basis to which we will reduce all degrees;
• Carefully perform transformations to get 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 this does not change the meaning;
• Cross out the bases of the 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 that will be told to you on this topic is just specific tricks and tricks to simplify and speed up the transformation. Here's one of those tricks we'll talk about now. :)

rationalization method

Consider another batch of inequalities:

\[\begin(align) & ((\text( )\!\!\pi\!\!\text( ))^(x+7)) \gt ((\text( )\!\!\pi \!\!\text( ))^(((x)^(2))-3x+2)); \\ & ((\left(2\sqrt(3)-3 \right))^(((x)^(2))-2x)) \lt 1; \\ & ((\left(\frac(1)(3) \right))^(((x)^(2))+2x)) \gt ((\left(\frac(1)(9) \right))^(16-x)); \\ & ((\left(3-2\sqrt(2) \right))^(3x-((x)^(2)))) \lt 1. \\\end(align)\]

Well, what is so special about them? They are also lightweight. Although, stop! Is pi raised to a power? What kind of nonsense?

And how to raise the number $2\sqrt(3)-3$ to a power? Or $3-2\sqrt(2)$? The compilers of the problems obviously drank too much "Hawthorn" before sitting down to work. :)

In fact, there is nothing wrong with 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 for one. The number π is positive - we already know this. The numbers $2\sqrt(3)-3$ and $3-2\sqrt(2)$ are also positive - this is easy to see if we compare them with zero.

It turns out that all these “terrifying” inequalities are no different from the simple ones discussed above? And they do it the same way? Yes, absolutely right. However, using their example, I would like to consider one trick that saves a lot of time on independent work and exams. Let's talk about 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 next game? Nothing like this! But this simple fact, written literally in one line, will greatly simplify our work. Take a look:

\[\begin(matrix) ((\text( )\!\!\pi\!\!\text( ))^(x+7)) \gt ((\text( )\!\!\pi\ !\!\text( ))^(((x)^(2))-3x+2)) \\ \Downarrow \\ \left(x+7-\left(((x)^(2)) -3x+2 \right) \right)\cdot \left(\text( )\!\!\pi\!\!\text( )-1 \right) \gt 0 \\\end(matrix)\]

Here 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 fucking multiplier \[\left(\text( )\!\!\pi\!\!\text( )-1 \right)\]? We don't know what the exact value of pi 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 doesn’t bother us much - it’s only important for us to understand that in any case $\text( )\!\!\pi\!\!\text( )-1 \gt 2$, t .e. 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 point, we had to divide by minus one, and the inequality sign changed. At the end, I expanded the square trinomial according to the Vieta theorem - it is obvious that the roots are equal to $((x)_(1))=5$ and $((x)_(2))=-1$. Then everything is solved by the classical method of intervals:

All points are punctured because the original inequality is strict. We are interested in the area 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 is simple here, because there is a unit on the right. And we remember that a unit is any number raised to the power of zero. Even if this number is an irrational expression, standing at the base on the left:

\[\begin(align) & ((\left(2\sqrt(3)-3 \right))^(((x)^(2))-2x)) \lt 1=((\left(2 \sqrt(3)-3\right))^(0)); \\ & ((\left(2\sqrt(3)-3 \right))^(((x)^(2))-2x)) \lt ((\left(2\sqrt(3)-3 \right))^(0)); \\\end(align)\]

So 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)\ ]

It remains only to deal with the signs. The multiplier $2\left(\sqrt(3)-2 \right)$ does not contain the variable $x$ - it's just a constant, and we need to figure 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 will change 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 quite 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 $f\left(x \right)=x\left(x-2 \right)$:

We are interested in the intervals marked with a plus sign. It remains only 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 quite obvious here: the bases are 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, in the process of transformations, we had to multiply by a negative number, so the inequality sign changed. At the very end, I again applied Vieta's theorem to factorize a square trinomial. As a result, the answer will be the following: $x\in \left(-8;4 \right)$ - those who wish can verify this by drawing a number line, marking points and counting signs. In the meantime, 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, the base is again an irrational number, and the unit is again on the right. Therefore, we rewrite our exponential inequality as follows:

\[((\left(3-2\sqrt(2) \right))^(3x-((x)^(2)))) \lt ((\left(3-2\sqrt(2) \ right))^(0))\]

Let's rationalize:

\[\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 parts of the inequality can be divided:

\[\begin(matrix) \left(3x-((x)^(2))-0 \right)\cdot 2\left(1-\sqrt(2) \right) \lt 0 \\ \Downarrow \ \\end(matrix)\]

\[\begin(align) & 3x-((x)^(2))-0 \gt 0; \\ & 3x-((x)^(2)) \gt 0;\quad \left| \cdot \left(-1 \right) \right. \\ & ((x)^(2))-3x \lt 0; \\ & x\left(x-3 \right) \lt 0. \\\end(align)\]

Change to another base

A separate problem in solving exponential inequalities is the search for the “correct” basis. Unfortunately, at the first glance at the task, it is far from always obvious what to take as a basis, and what to do as the degree of this basis.

But do not worry: there is no magic and "secret" technologies 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, these are:

\[\begin(align) & ((2)^(\frac(x)(2))) \lt ((4)^(\frac(4)(x))); \\ & ((\left(\frac(1)(3) \right))^(\frac(3)(x)))\ge ((3)^(2+x)); \\ & ((\left(0,16 \right))^(1+2x))\cdot ((\left(6,25 \right))^(x))\ge 1; \\ & ((\left(\frac(27)(\sqrt(3)) \right))^(-x)) \lt ((9)^(4-2x))\cdot 81. \\\ end(align)\]

Difficult? Scary? Yes, it's easier than 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 the 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 understood correctly: I just applied the rationalization method described above. Now we need to work carefully: we got a fractional-rational inequality (this is one that has a variable in the denominator), so before equating something to zero, you need to reduce 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 $x=0$. In total, there are three points that should be marked on the number line (all points are punched out, because the inequality sign is strict). We get:

As you might guess, hatching marks the intervals at which the expression on the left takes negative values. Therefore, two intervals will go into the final answer at once:

The ends of the intervals are not included in the answer because the original inequality was strict. No further validation of this answer is required. In this regard, exponential inequalities are much simpler than logarithmic ones: no DPV, 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 like this:

\[\begin(align) & ((\left(((3)^(-1)) \right))^(\frac(3)(x)))\ge ((3)^(2+x ))\Rightarrow ((3)^(-\frac(3)(x)))\ge ((3)^(2+x)); \\ & \left(-\frac(3)(x)-\left(2+x \right) \right)\cdot \left(3-1 \right)\ge 0; \\ & \left(-\frac(3)(x)-2-x \right)\cdot 2\ge 0;\quad \left| :\left(-2\right)\right. \\ & \frac(3)(x)+2+x\le 0; \\ & \frac(((x)^(2))+2x+3)(x)\le 0. \\\end(align)\]

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 the deuce was reduced with a constant multiplier. This is exactly what you should do when making real calculations for independent and control work - you do not need to paint every action and transformation directly.

Next, the familiar method of intervals comes into play. Zeros of the numerator: but there are none. Because the discriminant will be negative. In turn, the denominator is set to zero only when $x=0$ — just like last time. Well, it is clear that the fraction will take positive values ​​to the right of $x=0$, and negative ones to the left. Since we are only interested in negative values, the final answer is $x\in \left(-\infty ;0 \right)$.

\[((\left(0,16 \right))^(1+2x))\cdot ((\left(6,25 \right))^(x))\ge 1\]

And what should be done with decimal fractions in exponential inequalities? That's right: get rid of them by converting them into ordinary ones. Here we are translating:

\[\begin(align) & 0,16=\frac(16)(100)=\frac(4)(25)\Rightarrow ((\left(0,16 \right))^(1+2x)) =((\left(\frac(4)(25) \right))^(1+2x)); \\ & 6,25=\frac(625)(100)=\frac(25)(4)\Rightarrow ((\left(6,25 \right))^(x))=((\left(\ frac(25)(4) \right))^(x)). \\\end(align)\]

Well, what did we get in the bases of exponential functions? And we got two mutually reciprocal 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 indicators add up, which happened in the second line. In addition, we have represented the unit on the right, also as a power in base 4/25. It remains only 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 divided 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 ​​a solution here is also clear: all the exponential functions that make up the inequality must be reduced to the base "3". But for this you have to tinker a little with roots and degrees:

\[\begin(align) & \frac(27)(\sqrt(3))=\frac(((3)^(3)))(((3)^(\frac(1)(3)) ))=((3)^(3-\frac(1)(3)))=((3)^(\frac(8)(3))); \\ & 9=((3)^(2));\quad 81=((3)^(4)). \\\end(align)\]

Given these facts, 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 calculations: before doing something with 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 left or right left multipliers, extra constants, etc., no rationalization and "crossing out" of the grounds can be performed! Countless tasks have been done wrong due to a misunderstanding of this simple fact. I myself constantly observe this problem with my students when we are just starting to analyze exponential and logarithmic inequalities.

But back to our task. Let's try this time to do without rationalization. We recall: 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: $x\in \left(-\infty ;3 \right)$.

Highlighting a stable expression and replacing a variable

In conclusion, I propose to solve 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 bracketed. 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 with the very first line. Let's 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 side can be rewrite:

Note that there are no other exponential functions except for $((5)^(x+1))$ in the inequality. And in general, the variable $x$ does not occur 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 whole 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 decision on real control and independent work.

Well, let's try something more difficult. For example, here is an inequality:

\[((25)^(x+1,5))-((5)^(2x+2)) \gt 2500\]

What is the problem here? First of all, the bases of the exponential functions on the left are different: 5 and 25. However, 25 \u003d 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 is easily reduced to the second - it is enough just to expand the exponent. Now we can safely introduce a new variable: $((5)^(2x+2))=t$, and the whole inequality will be rewritten like this:

\[\begin(align) & 5t-t\ge 2500; \\ & 4t\ge 2500; \\ & t\ge 625=((5)^(4)); \\ & ((5)^(2x+2))\ge ((5)^(4)); \\ & 2x+2\ge 4; \\ & 2x\ge 2; \\ & x\ge 1. \\\end(align)\]

Again, no problem! Final answer: $x\in \left[ 1;+\infty \right)$. Moving 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 degree. 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 have taken the first step - everything has led to the same foundation. Now we need to highlight the 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 find out 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 with the three (numbers 9, 27, 81 and 243 are its powers), and with the seven (numbers 49 and 343 would also be nice to remember). Well, the five also have “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, all these numbers, if desired, can be restored in the mind, simply by successively multiplying them 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 there. And in this sense, these problems are more complex than the "classical" inequalities, which are solved by the interval method.

In this lesson, we will consider the solution of more complex exponential equations, recall the main theoretical provisions regarding the exponential function.

1. Definition and properties of an exponential function, a technique for solving the simplest exponential equations

Recall the definition and main properties of an exponential function. It is on the properties that the solution of all exponential equations and inequalities is based.

Exponential function is a function of the form , where the base is the degree and Here x is an independent variable, an argument; y - dependent variable, function.

Rice. 1. Graph of the exponential function

The graph shows an increasing and decreasing exponent, illustrating the exponential function at 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 as , decreases as .

A monotonic function takes each of its values ​​with a single value of the argument.

When the argument increases from minus to plus infinity, the function increases from zero, inclusive, to plus infinity. On the contrary, when the argument increases from minus to plus infinity, the function decreases from infinity to zero, inclusive.

2. Solution of typical exponential equations

Recall how to solve the simplest exponential equations. Their solution is based on the monotonicity of the exponential function. Almost all complex exponential equations are reduced to such equations.

The equality of exponents with equal bases is due to the property of the exponential function, namely its monotonicity.

Solution Method:

Equalize the bases of the degrees;

Equate exponents.

Let's move on to more complex exponential equations, our goal is to reduce each of them to the simplest.

Let's get rid of the root on the left side and reduce the degrees to the same base:

In order to reduce a complex exponential equation to a simple one, a change of variables is often used.

Let's use the degree property:

We introduce a replacement. Let , then . With this replacement, it is obvious that y takes strictly positive values. We get:

We multiply the resulting equation by two and transfer all the terms to the left side:

The first root does not satisfy the interval of y values, we discard it. We get:

Let's bring the degrees to the same indicator:

We introduce a replacement:

Let then . With this replacement, it is obvious that y takes strictly positive values. We get:

We know how to solve similar quadratic equations, we write out the answer:

To make sure that the roots are found correctly, you can check according to the Vieta theorem, that is, find the sum of the roots and their product and check with the corresponding coefficients of the equation.

We get:

3. Technique for solving homogeneous exponential equations of the second degree

Let us study the following important type of exponential equations:

Equations of this type are called homogeneous of the second degree with respect to the functions f and g. On its left side there is a square trinomial with respect to f with parameter g or a square trinomial with respect to g with parameter f.

Solution Method:

This equation can be solved as a quadratic one, but it is easier to do it the other way around. Two cases should be considered:

In the first case, we get

In the second case, we have the right to divide by the highest degree and we get:

You should introduce a change of variables , we get a quadratic equation for y:

Note that the functions f and g can be arbitrary, but we are interested in the case when these are exponential functions.

4. Examples of solving homogeneous equations

Let's move all the terms to the left side of the equation:

Since the exponential functions acquire strictly positive values, we have the right to immediately divide the equation by , without considering the case when:

We get:

We introduce a replacement: (according to the properties of the exponential function)

We got a quadratic equation:

We determine the roots according to the Vieta theorem:

The first root does not satisfy the interval of y values, we discard it, we get:

Let's use the properties of the degree and reduce all degrees to simple bases:

It is easy to notice the functions f and g:

Since the exponential functions acquire strictly positive values, we have the right to immediately divide the equation by , without considering the case when .