**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:

$$x \in \left<- \infty, 0\right> \cup \left[1, \infty\right>.$$