Now that we’ve learned definition and properties of exponentiation, it is natural to start wondering how to solve equations containing exponents. They are called **exponential equations**.

While solving exponential equations, we”ll use the following fact:

**$$a^x=a^y \Rightarrow x=y,$$ **

where $a$ is **the base** of the exponentiation and $x, y$ **the exponents**.

Let’s take a look at examples.

**Example 1: **

a) Solve $5^x=5^7$.

b) Solve $6^{1-x}=6^5$.

**Solution: **

a) The bases ($a=5$) are the same, so we immediately conclude that the exponents are also the same, i.e. $x=7$.

b) Here we also have the same bases ($a=6$), so the exponents also must be the same, i.e. $1-x=5$, from which it follows $x=6$.

Sometimes we will see equations with **different bases**. In that case, we need to **convert** one side (or both) of the equation, in order to use the mentioned rule.

**Example 2: **

a) Solve $4^x=16$.

b) Solve $3^{4x-1}=27$.

c) Solve $3^{x^2+3x}=81$.

**Solution: **

a) Noticing that $16=4^2$, we have $$4^x=4^2.$$ Now the bases are equal ($a=4$), so we conclude $x=2$.

b) Noticing that $27=3^3$, we have $$3^{4x-1}=3^3.$$ Therefore,

$$4x-1=3$$

$$4x=3+1$$

$$4x=4$$

$$x=1$$

We can always check if the number we’ve got is the solution.

**Check:**

$$3^{4 \cdot 1 – 1}= 3^{3}=27$$

Therefore, $x=1$ really is the solution.

c) Noticing that $81=3^4$, we have $$3^{x^2+3x}=3^4.$$

$$x^2+3x=4$$

$$x^2+3x-4=0$$

$$x_{1,2}=\frac{-3 \pm \sqrt{9+16}}{2}$$

$$x_{1,2}=\frac{-3 \pm 5}{2}$$

$$x_{1}=1, x_{2}=-4$$

**Check:**

For $x_{1}=1$: $3^{1^2+3 \cdot 1}=3^4=81$

For $x_{2}=-4$: $3^{(-4)^2+3 \cdot (-4)}=3^{16-12}=3^4=81$

Therefore, this equation has two solutions: $x_{1}=1, x_{2}=-4$.

In some equations we will need **negative exponents**:

* Example 3: *Solve $4^{5x+2}=\frac{1}{64}$.

**Solution:**

Noticing that $\frac{1}{64}=\frac{1}{4^3}=4^{-3}$, we have $$4^{5x+2}=4^{-3}$$

$$5x+2=-3$$

$$5x=-3-2$$

$$5x=-5$$

$$x=-\frac{5}{5}$$

$$x=-1$$

**Check:**

$$4^{5 \cdot (-1)+2}=4^{-3}=\frac{1}{64}$$

Therefore, $x=-1$ is the solution.

* Example 4: * Solve $4^{x}=-3$.

**Solution:**

Think about it; which exponent of positive number $4$ can give a negative number? For example, if we tried $x=0$, the result would be $4^0=1$. If we tried a negative number (for example $x=-3$), the result would be $4^{-3}=\frac{1}{64}$. If we tried any positive number $x$, the result would also be some positive number. Therefore, in all possible cases, **the result of exponentiation is always a positive number! **We can conclude that this equation has **no solutions**.

* Example 5: * Solve $2^x=31$.

**Solution:**

Here we have different bases and we can’t rearrange the equation to get the same bases, since $31$ is not the power of $2$. In this and similar examples, we”ll use a rule:

$$log_{a}x^{n}=nlog_{a}x.$$

Therefore,

$$log_{2}2^x=log_{2}31$$

$$x \cdot log_{2}2=log_{2}31$$

Using **$log_{a}a=1$**, we have

$$x=log_{2}31$$

Using the change of base rule, we have

$$x=\frac{ln31}{ln2} \approx 4.95.$$

**Note: **We could immediately use** ln** instead of

*$log_{2}$*; we would get the same answer.

* Example 6: *Solve $5(2^{x-4})=203$.

**Solution:**

We will first divide the whole equation by $5$:

$$2^{x-4}=\frac{203}{5}$$

Since $\frac{203}{5}$ is not the power of $2$, we need to use a log rule:

$$ln(2^{x-4})=ln\left(\frac{203}{5}\right)$$

$$(x-4)ln2=ln\left(\frac{203}{5}\right)$$

$$x-4=\frac{ln\left(\frac{203}{5}\right)}{ln2}$$

$$x=\frac{ln\left(\frac{203}{5}\right)}{ln2}+4 \approx 9.34$$

* Example 7: *Solve $4^{6-9x}=\frac{1}{8^{x-3}}$.

**Solution:**

Noticing that $4=2^2$ and $8=2^3$, we have

$$(2^2)^{6-9x}=\frac{1}{(2^3)^{x-3}}$$

$$2^{2 \cdot (6-9x)}=\frac{1}{2^{3 \cdot (x-3)}}$$

We don’t want to have a fraction on the right side of the equation, so we apply the rule:

**$$\frac{1}{a^n}=a^{-n}.$$**

Using this gives

$$2^{2 \cdot (6-9x)}=2^{-3 \cdot (x-3)}$$

Since the bases are equal, we have

$$2 \cdot (6-9x)=-3 \cdot (x-3)$$

$$12-18x=-3x+9$$

$$-18x+3x=9-12$$

$$-15x=-3$$

$$x=\frac{-3}{-15}$$

$$x=\frac{1}{5}$$

**Check:**

Left side: $4^{6-9 \cdot \frac{1}{5}}=4^{6-\frac{9}{5}}=4^{\frac{21}{5}} \approx 337.79$

Right side: $\frac{1}{8^{\frac{1}{5}-3}}=\frac{1}{8^{-\frac{14}{5}}}=8^{\frac{14}{5}} \approx 337.79$

Therefore, $x=\frac{1}{5}$ is the solution.

* Example 8: *Solve $3e^x+6=486$.

**Solution:**

First we need to isolate the variable, i.e. $e^x$:

$$3e^x=486-6$$

$$3e^x=480$$

Dividing by $3$,

$$e^x=160$$

Since the bases aren’t equal, we will take the *ln *of both sides:

$$ln(e^x)=ln(160)$$

$$x \cdot ln(e)=ln(160)$$

$$x=ln(160) \approx 5.075$$