Exponential inequalities are inequalities in which one or both sides contain a variable exponent. While solving exponential inequalities, we must keep in mind these facts:
1) If $a>1$ and $x<y$, then $a^x<a^y$.
If $a>1$ and $a^x<a^y$, then $x<y$.
Precisely, the exponential function $f(x)=a^x$ is monotonically increasing for $a>1$.
2) If $0<a<1$ and $x<y$, then $a^x>a^y$.
If $0<a<1$ and $a^x>a^y$, then $x<y$.
Precisely, the exponential function $f(x)=a^x$ is monotonically decreasing for $0<a<1$.
Exponential inequalities – same bases
Example 1:
$$2^{3x+4}>2^{2x}$$
Solution:
The base is $a=2$, which is greater than $1$. This implies $3x+4>2x$.
$$3x-2x>-4$$
$$x>-4$$
Therefore, the solutions are $x>-4$, i.e. $x \in \left<-4, \infty\right>$.
Example 2:
$$\left(\frac{1}{2}\right)^{5x}<\left(\frac{1}{2}\right)^{3x-4}$$
Solution:
The base is $a=\frac{1}{2}$, which is less than $1$. This implies $5x>3x-4$.
$$5x-3x>-4$$
$$2x>-4$$
Dividing by $2$,
$$x>-2$$
Therefore, the solutions are $x>-2$, i.e. $x \in \left<-2, \infty\right>$.
Exponential inequalities – different bases
Example 3:
$$8^{2x-3}>4^{4x}$$
Solution:
In this case, the bases are different but similar. Precisely, we know that $8=2^3$ and $4=2^2$.
$$\left(2^3\right)^{2x-3}>\left(2^2\right)^{4x}$$
$$2^{3 \cdot (2x-3)}>2^{2 \cdot 4x}$$
The base is $a=2>1$, so we conclude
$$3 \cdot (2x-3)>2 \cdot 4x$$
$$6x-9>8x$$
$$6x-8x>9$$
$$-2x>9$$
Dividing by $-2$,
$$x<-\frac{9}{2}$$
Therefore, the solutions are $x<-\frac{9}{2}$, i.e. $x \in \left<-\infty, -\frac{9}{2}\right>$.
Example 4:
$$5^{2x+1}<2^{x}$$
Solution:
The bases are different, so we need to take logarithm of both sides of the inequality:
$$log5^{2x+1}<log2^{x}$$
Now we apply the logarithm power rule $log_{a}x^y=y \cdot log_{a}x$:
$$(2x+1)log5<xlog2$$
Rearranging gives us
$$2xlog5+log5<xlog2$$
$$x(2log5-log2)<-log5$$
We can again apply the logarithm power rule:
$$x(log5^2-log2)<-log5$$
Using the logarithm quotient rule, we get
$$x \cdot log\frac{25}{2}< -log5$$
Dividing by $log\frac{25}{2}$,
$$x<-\frac{log5}{log\frac{25}{2}} \approx -0.637$$
Therefore, the solutions are $x<- \frac{log5}{log\frac{25}{2}} \approx -0.637$, i.e. $x \in \left<-\infty, -\frac{log5}{log\frac{25}{2}} \right>$.
Other examples
Example 5:
$$4^{2x+1}>4^x+3$$
Solution:
Rearranging gives us
$$4^{2x} \cdot 4>4^x+3$$
We use a substitution $4^x=t$:
$$4t^2>t+3$$
$$4t^2-t-3>0$$
Now we need to solve the related quadratic equation $4t^2-t-3=0$.
$$t_{1,2}=\frac{1 \pm \sqrt{1-4 \cdot 4 \cdot (-3)}}{8}$$
$$=\frac{1 \pm \sqrt{49}}{8}$$
$$=\frac{1 \pm 7}{8}$$
$$t_{1}=1, t_{2}=-\frac{3}{4}$$
This means that we need to solve inequalities $t<-\frac{3}{4}$ and $t>1$.
Since $t=4^x$, inequality $t<-\frac{3}{4}$ has no solutions, because the values of exponential function are always positive numbers.
$$t>1 \Rightarrow 4^x>1$$
$$4^x>4^0$$
$$x>0$$
Therefore, the solutions are $x>0$, i.e. $x \in \left<0, \infty\right>$.
Example 6:
$$\frac{2^x-2^{1-x}-1}{1-2^x}\leq0$$
Solution:
Think about it; when can a fraction be a nonnegative number? We have two cases:
1) $$2^x-2^{1-x}-1 \leq 0$$
$$1-2^x>0$$
2) $$2^x-2^{1-x}-1 \geq 0$$
$$1-2^x<0$$
Let’s solve the first case.
1) Rearranging of the first inequality gives us
$$2^x-2 \cdot 2^{-x}-1 \leq 0$$
We use a substitution $2^x=t$:
$$t-\frac{2}{t}-1 \leq 0$$
$$t^2-2-t \leq 0$$
We need to solve the related quadratic equation:
$$t^2-2-t=0$$
$$t_{1,2}=\frac{1 \pm \sqrt{9}}{2}=\frac{1 \pm 3}{2}$$
$$t_{1}=2, t_{2}=-1$$
This means that we need to solve inequalities $t \leq 2$ and $t \geq -1$.
$$t \leq 2 \Rightarrow 2^x \leq 2 \Rightarrow x \leq 1$$
$$t \geq -1 \Rightarrow 2^x \geq -1,$$
which is fulfilled for every $x \in \mathbf{R}$. Therefore, the solution of the first inequality is $x \leq 1$.
Now let’s take a look at the second inequality:
$$1-2^x>0 \Rightarrow 2^x<1 \Rightarrow x<0$$
Therefore, the solution of the first case is the intersection of the solutions of both inequalities: $x \leq 1 \cap x<0 = {x \in \mathbf{R}: x<0}$, i.e. the solutions in interval notation are $x \in \left<- \infty, 0\right>$.
Let’s solve the second case.
2) Similarly, the solutions of the inequality $2^x-2^{1-x}-1 \leq 0$ are $x \in \left[1, \infty\right>$ and the solutions of $1-2^x<0$ are $x \in \left<0, \infty\right>$. Therefore, the solution of the second case is the intersection of the solutions of both inequalities: $x \in \left[1, \infty\right>$.
Final solution is the union of the solutions of the first and the second case, i.e. the solution is:
$$x \in \left<- \infty, 0\right> \cup \left[1, \infty\right>.$$
