Exponential equations

An expo­nen­tial equa­tion is one in which the unknown is in the expo­nent, such as $4^{x} = (2\sqrt{2})^{8}$ or $\displaystyle 27^{x} = \frac{9}{3^{x}}$. You can­not solve these by trans­pos­ing as in a lin­ear equa­tion, because $x$ is not mul­ti­ply­ing or adding any­thing: it is a power. Instead you use Law 11: if $a^{m} = a^{n}$, then $m = n$. This les­son shows how to bring both sides to the same base and then com­pare expo­nents.

The method#

1. Write both sides as pow­ers of the same base, usu­ally a prime: $4 = 2^{2}$, $27 = 3^{3}$, $\displaystyle \frac{1}{32} = 2^{-5}$, $\sqrt{2} = 2^{1/2}$.
2. Use the laws to make each side a sin­gle power of that base.
3. Equate the expo­nents (Law 11) and solve the equa­tion that results, which is usu­ally lin­ear.
4. Check, and answer the ques­tion actu­ally asked.

Exam­ple 1#

If $4^{x} = (2\sqrt{2})^{8}$, find $x + 2$.

First write $2\sqrt{2}$ as a power of $2$: $2\sqrt{2} = 2^{1} \times 2^{1/2} = 2^{1 + 1/2} = 2^{3/2}$.

$$\begin{aligned} (2^{2})^{x} &= (2^{3/2})^{8} \\ 2^{2x} &= 2^{12} \\ 2x &= 12 \\ x &= 6 \end{aligned}$$

Here $\displaystyle \frac{3}{2} \times 8 = 12$. The ques­tion asked for $x + 2$, so the answer is $6 + 2 = 8$.

Exam­ple 2#

Solve $2^{5x} \div 2^{x} = \sqrt[5]{2^{20}}$.

Use Law 2 on the left and Law 13 on the right.

$$\displaystyle \begin{aligned} \frac{2^{5x}}{2^{x}} &= 2^{20/5} \\ 2^{5x - x} &= 2^{4} \\ 2^{4x} &= 2^{4} \\ 4x &= 4 \\ x &= 1 \end{aligned}$$

Exam­ple 3: two dif­fer­ent bases#

Solve $5^{x - 2} \times 3^{2x - 3} = 135$.

Here there are two primes. Break $135$ into primes: $135 = 5 \times 27 = 5^{1} \times 3^{3}$.

$$5^{x - 2} \times 3^{2x - 3} = 5^{1} \times 3^{3}$$

The pow­ers of $5$ must match and the pow­ers of $3$ must match:

$$x - 2 = 1 \quad \text{and} \quad 2x - 3 = 3$$

Both give $x = 3$. Good: the two con­di­tions agree, so $x = 3$.

Exam­ple 4: no need to find x#

If $3^{x} = 270$, find $3^{x - 3}$.

There is no need to find $x$. Use Law 2 back­wards: $\displaystyle 3^{x - 3} = \frac{3^{x}}{3^{3}}$.

$$\displaystyle 3^{x - 3} = \frac{3^{x}}{3^{3}} = \frac{270}{27} = 10$$

Exam­ple 5: a neg­a­tive answer#

If $\displaystyle 4^{2x} = \frac{1}{32}$, find $x$.

$$\displaystyle \begin{aligned} (2^{2})^{2x} &= \frac{1}{2^{5}} \\ 2^{4x} &= 2^{-5} \\ 4x &= -5 \\ x &= -\frac{5}{4} \end{aligned}$$

Exam­ple 6: a frac­tional base#

Solve $\displaystyle \left(\sqrt{\frac{3}{5}}\right)^{x + 1} = \frac{125}{27}$.

The base on the left is $\displaystyle \left(\frac{3}{5}\right)^{1/2}$. The right side is $\displaystyle \left(\frac{5}{3}\right)^{3}$, which by Law 8 is $\displaystyle \left(\frac{3}{5}\right)^{-3}$.

$$\displaystyle \begin{aligned} \left[\left(\frac{3}{5}\right)^{1/2}\right]^{x + 1} &= \left(\frac{3}{5}\right)^{-3} \\ \left(\frac{3}{5}\right)^{\frac{x + 1}{2}} &= \left(\frac{3}{5}\right)^{-3} \\ \frac{x + 1}{2} &= -3 \\ x + 1 &= -6 \\ x &= -7 \end{aligned}$$

Exam­ple 7: the unknown power appears twice#

Solve $25^{x - 1} = 5^{2x - 1} - 100$.

Because of the $-100$, the two sides can­not be made into sin­gle pow­ers. Instead write every­thing in terms of one quan­tity, $5^{2x}$:

$$\displaystyle 25^{x - 1} = 5^{2x - 2} = \frac{5^{2x}}{25}, \qquad 5^{2x - 1} = \frac{5^{2x}}{5}$$

So the equa­tion is

$$\displaystyle \begin{aligned} \frac{5^{2x}}{25} &= \frac{5^{2x}}{5} - 100 \\ \frac{5^{2x}}{25} - \frac{5^{2x}}{5} &= -100 \\ 5^{2x}\left(\frac{1}{25} - \frac{1}{5}\right) &= -100 \\ 5^{2x}\left(\frac{1 - 5}{25}\right) &= -100 \\ 5^{2x}\left(-\frac{4}{25}\right) &= -100 \\ 5^{2x} &= 625 = 5^{4} \\ 2x &= 4 \\ x &= 2 \end{aligned}$$

Check: $25^{1} = 25$ and $5^{3} - 100 = 125 - 100 = 25$. You can also work with $25^{x}$ in place of $5^{2x}$, since $5^{2x} = 25^{x}$; the steps are the same.

Exam­ple 8#

Solve $\displaystyle 27^{x} = \frac{9}{3^{x}}$.

Mul­ti­ply both sides by $3^{x}$, then use base $3$.

$$\displaystyle \begin{aligned} (3^{3})^{x} \cdot 3^{x} &= 3^{2} \\ 3^{3x} \cdot 3^{x} &= 3^{2} \\ 3^{4x} &= 3^{2} \\ 4x &= 2 \\ x &= \frac{1}{2} \end{aligned}$$

Prac­tice#

1. Solve $2^{x + 3} = 64$.
2. Solve $9^{x} = 27$.
3. Solve $8^{x - 1} = 4^{x + 1}$.
4. Solve $\displaystyle 3^{2x - 1} = \frac{1}{27}$.
5. Solve $\displaystyle \left(\frac{2}{5}\right)^{x + 2} = \frac{125}{8}$.
6. Solve $2^{x} \times 3^{x + 1} = 108$.
7. If $2^{x} = 48$, find $2^{x - 4}$.
8. Solve $(\sqrt[3]{3})^{x - 1} = 9$.
9. If $3^{x + 1} = 9^{x - 2}$, find the value of $2^{1 + x}$.
10. Solve $5^{x + 1} + 5^{x} = 750$.
11. Solve $4^{x} - 4^{x - 1} = 24$.

Answers#