Solving Exponential Equations

An exponential equation has the unknown in an exponent, like $5^x = 40$ or $2000(1.045)^n = 3000$. In Grade 11 you solved these by matching bases or by guess and check. Now you have logarithms, which can bring any exponent down to where you can solve for it. This page shows both methods and how to choose between them.

Key ideas

Method 1: a common base

If two powers of the same base are equal, their exponents are equal:

$$b^m = b^n \quad\Rightarrow\quad m = n \qquad (b \gt 0,\ b \ne 1)$$

This works when both sides can be written as powers of the same number. For example, $4^x = 8^{x - 1}$ becomes $2^{2x} = 2^{3x - 3}$, so $2x = 3x - 3$ and $x = 3$.

Method 2: take logarithms

When the numbers aren’t powers of a common base, take the log of both sides. The power law brings the exponent down:

$$5^x = 40 \quad\Rightarrow\quad \log\left(5^x\right) = \log 40 \quad\Rightarrow\quad x\log 5 = \log 40 \quad\Rightarrow\quad x = \frac{\log 40}{\log 5}$$

Base $10$ is the usual choice because it’s on your calculator, but any base works. (Rewriting $5^x = 40$ in log form, $x = \log_5 40$, gives the same answer after change of base.)

Before taking logs, isolate the power. In $2000(1.045)^n = 3000$, divide by $2000$ first.

Sums of powers: factor first

You can’t take the log of each term in a sum. But if the terms share a power, factor it out:

$$2^{x + 3} - 2^x = 2^x \cdot 2^3 - 2^x = 2^x(8 - 1) = 7\left(2^x\right)$$

Quadratic-type equations

Equations like $9^x - 4\left(3^x\right) + 3 = 0$ are quadratics in disguise, because $9^x = \left(3^2\right)^x = \left(3^x\right)^2$. Let $u = 3^x$, solve the quadratic for $u$, then solve $3^x = u$. Remember that $3^x$ is always positive, so a negative or zero value of $u$ gives no solution.

Choosing a method

The equation looks like…	Use
both sides are powers of the same number ($8$ and $32$, or $9$ and $\tfrac{1}{27}$)	common base; gives an exact answer
the numbers don’t share a base ($5^x = 40$, growth and interest problems)	take logarithms
a sum or difference of powers of the same base	factor out the common power first
one power is the square of another ($4^x$ and $2^x$)	substitute $u$, then solve a quadratic

Worked examples

Example 1: A common base

Solve.

• (a) $8^{x + 1} = 4^{2x}$
• (b) $9^{x - 1} = \dfrac{1}{27}$

Solution.

(a) Both $8$ and $4$ are powers of $2$:

$$\begin{aligned} \left(2^3\right)^{x + 1} &= \left(2^2\right)^{2x} \\ 2^{3x + 3} &= 2^{4x} \\ 3x + 3 &= 4x && \text{equal bases, so equal exponents} \\ x &= 3 \end{aligned}$$

Check: $8^4 = 4096$ and $4^6 = 4096$. ✓

(b) Both $9$ and $27$ are powers of $3$, and $\dfrac{1}{27} = 3^{-3}$:

$$3^{2(x - 1)} = 3^{-3} \quad\Rightarrow\quad 2x - 2 = -3 \quad\Rightarrow\quad x = -\tfrac{1}{2}$$

Check: $9^{-\frac{3}{2}} = \dfrac{1}{\left(\sqrt{9}\right)^3} = \dfrac{1}{27}$. ✓

Example 2: Taking logarithms

(a) Solve $5^x = 40$, to three decimal places.

(b) Shreya invests $2000 at $4.5\%$ per year, compounded annually. How many years until her investment is worth at least $3000?

Solution.

(a) $40$ isn’t a power of $5$, so take logs:

$$x\log 5 = \log 40 \quad\Rightarrow\quad x = \frac{\log 40}{\log 5} \approx 2.292$$

Check: $5^{2.292} \approx 40.0$. ✓

(b) The compound interest model is $A = 2000(1.045)^n$. Set $A = 3000$ and isolate the power before taking logs:

$$\begin{aligned} 2000(1.045)^n &= 3000 \\ 1.045^n &= 1.5 && \text{divide by } 2000 \\ n\log 1.045 &= \log 1.5 && \text{take logs} \\ n &= \frac{\log 1.5}{\log 1.045} \approx 9.21 \end{aligned}$$

Interest is added only at the end of each year, so after $9$ years it isn’t quite there yet ($2000(1.045)^9 \approx$ $2972.19), and after $10$ years it is (about $3105.94). It takes $10$ years.

There’s no common base for $1.045$ and $1.5$, which is why logs are the right method here.

Example 3: Different bases on each side

Solve $3^{2x - 1} = 7^x$. Give an exact answer and a decimal to three places.

Solution. Take logs of both sides, and keep the brackets around the exponent:

$$\begin{aligned} (2x - 1)\log 3 &= x\log 7 \\ 2x\log 3 - \log 3 &= x\log 7 && \text{expand} \\ 2x\log 3 - x\log 7 &= \log 3 && \text{collect } x \text{ terms} \\ x(2\log 3 - \log 7) &= \log 3 && \text{factor out } x \\ x &= \frac{\log 3}{2\log 3 - \log 7} \end{aligned}$$

The denominator simplifies to $\log 9 - \log 7 = \log \frac{9}{7}$, so $x = \dfrac{\log 3}{\log \frac{9}{7}} \approx 4.371$.

Check: $3^{2(4.371) - 1} \approx 4940$ and $7^{4.371} \approx 4940$. ✓

Example 4: Factoring and a hidden quadratic

Solve.

• (a) $3^{x + 2} - 3^x = 72$
• (b) $9^x - 4\left(3^x\right) + 3 = 0$

Solution.

(a) Factor out $3^x$, using $3^{x + 2} = 3^x \cdot 9$:

$$3^x(9 - 1) = 72 \quad\Rightarrow\quad 3^x = 9 \quad\Rightarrow\quad x = 2$$

Check: $3^4 - 3^2 = 81 - 9 = 72$. ✓ Factoring turned it into a common-base equation.

(b) Since $9^x = \left(3^x\right)^2$, let $u = 3^x$:

$$u^2 - 4u + 3 = 0 \quad\Rightarrow\quad (u - 1)(u - 3) = 0 \quad\Rightarrow\quad u = 1 \ \text{ or } \ u = 3$$

Now go back to $x$: $3^x = 1$ gives $x = 0$, and $3^x = 3$ gives $x = 1$.

Check $x = 1$: $9 - 4(3) + 3 = 0$. ✓ Check $x = 0$: $1 - 4 + 3 = 0$. ✓

Common mistakes

Taking the log of each term in a sum. $\log\left(3^{x + 2} - 3^x\right)$ is not $\log\left(3^{x + 2}\right) - \log\left(3^x\right)$. Factor out the common power first, as in Example 4(a).

Multiplying the coefficient into the base. $2000(1.045)^n$ is not $(2090)^n$. The exponent only applies to $1.045$. Divide by $2000$ before you do anything else.

Losing the brackets on the exponent. $\log\left(3^{2x - 1}\right) = (2x - 1)\log 3$, not $2x - 1\log 3$. Without brackets, only the $1$ gets multiplied.

Keeping an impossible value of $u$. In a quadratic-type equation, $2^x = -3$ has no solution, because a power of $2$ is always positive. Reject it.

Simplifying a quotient of logs into one log. $\dfrac{\log 40}{\log 5}$ is about $2.292$. It is not $\log 8$ (about $0.903$). Divide the two decimal values.

Rounding too early. Keep the full calculator values until the last step. In Example 2(b), rounding $\log 1.045$ to $0.02$ gives $n \approx 8.80$, and the wrong answer of $9$ years.

Practice

1. (Warm-up) Solve using a common base.

• (a) $2^x = 32$
• (b) $27^x = 9$
• (c) $4^{x - 1} = 64$

Solution

(a) $2^x = 2^5$, so $x = 5$.

(b) $3^{3x} = 3^2$, so $3x = 2$ and $x = \tfrac{2}{3}$.

(c) $4^{x - 1} = 4^3$, so $x - 1 = 3$ and $x = 4$.

2. (Warm-up) Solve $6^x = 15$, to three decimal places.

Solution

$$x = \frac{\log 15}{\log 6} \approx 1.511$$

Check: $6^{1.511} \approx 15.0$. ✓

3. (Core) Solve $\left(\tfrac{1}{2}\right)^x = 8^{x + 4}$.

Solution

Write both sides as powers of $2$: $\left(\tfrac{1}{2}\right)^x = 2^{-x}$ and $8^{x + 4} = 2^{3(x + 4)}$.

$$-x = 3x + 12 \quad\Rightarrow\quad -4x = 12 \quad\Rightarrow\quad x = -3$$

Check: $\left(\tfrac{1}{2}\right)^{-3} = 8$ and $8^1 = 8$. ✓

4. (Core) Solve $12(1.5)^x = 90$, to three decimal places.

Solution

$$1.5^x = 7.5 \quad\Rightarrow\quad x = \frac{\log 7.5}{\log 1.5} \approx 4.969$$

5. (Core) Solve $5^{x + 1} = 3^{2x}$. Give an exact answer and a decimal to three places.

Solution

$$\begin{aligned} (x + 1)\log 5 &= 2x\log 3 \\ x\log 5 + \log 5 &= 2x\log 3 \\ \log 5 &= x(2\log 3 - \log 5) \\ x &= \frac{\log 5}{2\log 3 - \log 5} = \frac{\log 5}{\log 1.8} \approx 2.738 \end{aligned}$$

(The denominator is $\log 9 - \log 5 = \log \tfrac{9}{5} = \log 1.8$.) Check: $5^{3.738} \approx 410$ and $3^{5.476} \approx 410$. ✓

6. (Core) Solve $2^{x + 3} + 2^x = 72$.

Solution

Factor out $2^x$: $2^x(8 + 1) = 72$, so $2^x = 8$ and $x = 3$.

Check: $2^6 + 2^3 = 64 + 8 = 72$. ✓

7. (Core) Liam deposits $5000 in an account paying $6\%$ per year, compounded monthly. How many months until the account holds at least $8000? About how many years is that?

Solution

The monthly rate is $\dfrac{0.06}{12} = 0.005$, so $A = 5000(1.005)^n$ with $n$ in months.

$$5000(1.005)^n = 8000 \quad\Rightarrow\quad 1.005^n = 1.6 \quad\Rightarrow\quad n = \frac{\log 1.6}{\log 1.005} \approx 94.24$$

After $94$ months the balance is about $7990.61 (not enough), and after $95$ months it’s about $8030.56. So it takes $95$ months, which is $7$ years and $11$ months.

8. (Challenge) Solve $4^x - 2^{x + 1} - 8 = 0$.

Solution

$4^x = \left(2^x\right)^2$ and $2^{x + 1} = 2\left(2^x\right)$. Let $u = 2^x$:

$$u^2 - 2u - 8 = 0 \quad\Rightarrow\quad (u - 4)(u + 2) = 0 \quad\Rightarrow\quad u = 4 \ \text{ or } \ u = -2$$

$2^x = 4$ gives $x = 2$. $2^x = -2$ has no solution, since $2^x \gt 0$. So $x = 2$.

Check: $16 - 8 - 8 = 0$. ✓

9. (Challenge) Solve $25^x - 7\left(5^x\right) + 10 = 0$. Give exact answers, then decimals where needed.

Solution

Let $u = 5^x$, so $25^x = u^2$:

$$u^2 - 7u + 10 = 0 \quad\Rightarrow\quad (u - 2)(u - 5) = 0 \quad\Rightarrow\quad u = 2 \ \text{ or } \ u = 5$$

$5^x = 5$ gives $x = 1$. $5^x = 2$ gives $x = \dfrac{\log 2}{\log 5} \approx 0.431$.

Check $x = 1$: $25 - 35 + 10 = 0$. ✓ Check $x \approx 0.431$: $5^x = 2$ and $25^x = 4$, so $4 - 14 + 10 = 0$. ✓