Solving Logarithmic Equations

A logarithmic equation has the unknown inside a logarithm, like $\log_4(3x - 2) = 2$. The main move is simple: switch to exponential form, and the log disappears. The catch is that logs only accept positive inputs, so some answers that come out of the algebra don’t actually work. Checking every answer is part of the method, not an extra.

Key ideas

Rewrite in exponential form

If one side is a single log and the other side is a number:

$$\log_b A = c \quad\Longleftrightarrow\quad A = b^c$$

For example, $\log_2(x + 1) = 3$ means $x + 1 = 2^3 = 8$, so $x = 7$.

Log equals log

A logarithmic function never takes the same value twice (it’s always increasing or always decreasing), so if two logs with the same base are equal, their arguments are equal:

$$\log_b A = \log_b B \quad\Rightarrow\quad A = B$$

Combine first

If there are several logs on one side, use the laws of logarithms to write them as a single log, then use one of the two moves above.

Check for extraneous roots

Every log in the original equation needs a positive argument. An answer that makes any argument zero or negative is extraneous, and you reject it.

Why do extra answers appear? Combining logs can widen the domain. In $\log_2 x + \log_2(x - 2)$, both $x$ and $x - 2$ must be positive, so $x \gt 2$. But the combined expression $\log_2\big(x(x - 2)\big)$ also works for negative $x$, since then $x(x - 2)$ is positive too. Solving the combined equation can pick up an answer that the original equation never had.

$\log_2 x + \log_2(x - 2) = 3$ has one solution, $x = 4$. The combined form $\log_2\big(x(x - 2)\big) = 3$ also gives $x = -2$, which is extraneous.

A negative answer isn’t automatically wrong, though. For $\log_2(x + 5) = 1$, the answer is $x = -3$, and it’s fine: $x + 5 = 2 \gt 0$. Check the arguments, not the sign of $x$.

Worked examples

Example 1: Rewriting in exponential form

Solve.

• (a) $\log_4(3x - 2) = 2$
• (b) $\log(2x + 4) = 2$

Solution.

(a)

$$3x - 2 = 4^2 = 16 \quad\Rightarrow\quad 3x = 18 \quad\Rightarrow\quad x = 6$$

Check: $3(6) - 2 = 16 \gt 0$, and $\log_4 16 = 2$. ✓

(b) The base is $10$:

$$2x + 4 = 10^2 = 100 \quad\Rightarrow\quad 2x = 96 \quad\Rightarrow\quad x = 48$$

Check: $\log(2(48) + 4) = \log 100 = 2$. ✓

Example 2: Log equals log

Solve.

• (a) $\log_3(2x + 1) = \log_3(x + 7)$
• (b) $2\log x = \log 36$

Solution.

(a) The logs have the same base, so the arguments are equal:

$$2x + 1 = x + 7 \quad\Rightarrow\quad x = 6$$

Check: $2(6) + 1 = 13$ and $6 + 7 = 13$, both positive. ✓

(b) Use the power law on the left: $\log\left(x^2\right) = \log 36$, so $x^2 = 36$ and $x = 6$ or $x = -6$.

Now check in the original equation. $x = 6$ works: $2\log 6 = \log 36$. ✓ But $x = -6$ makes $\log x = \log(-6)$, which is undefined. Reject it. The only solution is $x = 6$.

Example 3: Combining with the product law

Solve $\log_2 x + \log_2(x - 2) = 3$.

Solution. Combine into a single log, then rewrite in exponential form:

$$\begin{aligned} \log_2\big(x(x - 2)\big) &= 3 && \text{product law} \\ x(x - 2) &= 2^3 && \text{exponential form} \\ x^2 - 2x - 8 &= 0 \\ (x - 4)(x + 2) &= 0 \end{aligned}$$

So $x = 4$ or $x = -2$.

• $x = 4$: $\log_2 4 + \log_2 2 = 2 + 1 = 3$. ✓
• $x = -2$: $\log_2(-2)$ is undefined. Reject.

The solution is $x = 4$. The graph above shows why only one answer survives.

Example 4: Combining with the quotient law

Solve $\log_2(x + 6) - \log_2(x - 1) = 3$.

Solution.

$$\begin{aligned} \log_2 \frac{x + 6}{x - 1} &= 3 && \text{quotient law} \\ \frac{x + 6}{x - 1} &= 8 && \text{exponential form} \\ x + 6 &= 8(x - 1) && \text{multiply by } x - 1 \\ x + 6 &= 8x - 8 \\ 14 &= 7x \\ x &= 2 \end{aligned}$$

Check: $x + 6 = 8 \gt 0$ and $x - 1 = 1 \gt 0$, and $\log_2 8 - \log_2 1 = 3 - 0 = 3$. ✓

Common mistakes

Skipping the check. An answer can come out of perfectly correct algebra and still be extraneous. Substitute each answer into every log of the original equation, and make sure each argument is positive.

Rejecting an answer just because it’s negative. What matters is the arguments. In $\log_2(x + 5) = 1$, the answer $x = -3$ gives $\log_2 2 = 1$, which is fine.

“Cancelling” the logs when the sides aren’t single logs. $\log x + \log 2 = \log 10$ does not mean $x + 2 = 10$. Combine first: $\log(2x) = \log 10$, so $2x = 10$ and $x = 5$.

Rewriting in exponential form incorrectly. $\log_3 A = 2$ means $A = 3^2 = 9$. It doesn’t mean $A = 2^3$ or $A = 3 \times 2$. The base of the log becomes the base of the power.

Combining logs with the wrong law. $\log x + \log(x - 2)$ is $\log\big(x(x - 2)\big)$, a product. It is not $\log(2x - 2)$, which would mean adding the arguments.

Practice

1. (Warm-up) Solve.

• (a) $\log_2 x = 6$
• (b) $\log_5(x + 3) = 2$

Solution

(a) $x = 2^6 = 64$

(b) $x + 3 = 5^2 = 25$, so $x = 22$.

2. (Warm-up) Solve $\log(4x) = 3$.

Solution

$4x = 10^3 = 1000$, so $x = 250$.

3. (Warm-up) Solve $\log_7(3x - 1) = \log_7(x + 9)$.

Solution

$3x - 1 = x + 9$, so $2x = 10$ and $x = 5$. Check: both arguments equal $14 \gt 0$. ✓

4. (Core) Solve $\log_3(x + 2) + \log_3 4 = 2$.

Solution

$$\log_3\big(4(x + 2)\big) = 2 \quad\Rightarrow\quad 4(x + 2) = 9 \quad\Rightarrow\quad x + 2 = \tfrac{9}{4} \quad\Rightarrow\quad x = \tfrac{1}{4}$$

Check: $x + 2 = \tfrac{9}{4} \gt 0$, and $\log_3 \tfrac{9}{4} + \log_3 4 = \log_3 9 = 2$. ✓

5. (Core) Solve $\log x + \log(x + 21) = 2$.

Solution

$$\begin{aligned} \log\big(x(x + 21)\big) &= 2 \\ x^2 + 21x &= 100 \\ x^2 + 21x - 100 &= 0 \\ (x + 25)(x - 4) &= 0 \end{aligned}$$

$x = -25$ makes $\log x$ undefined, so reject it. $x = 4$: $\log 4 + \log 25 = \log 100 = 2$. ✓

The solution is $x = 4$.

6. (Core) Solve $\log_2(x + 3) - \log_2(x - 4) = 3$.

Solution

$$\frac{x + 3}{x - 4} = 2^3 = 8 \quad\Rightarrow\quad x + 3 = 8x - 32 \quad\Rightarrow\quad 35 = 7x \quad\Rightarrow\quad x = 5$$

Check: $x + 3 = 8$ and $x - 4 = 1$, both positive, and $\log_2 8 - \log_2 1 = 3$. ✓

7. (Core) Solve $2\log_5 x - \log_5 4 = 2$.

Solution

$$\log_5 \frac{x^2}{4} = 2 \quad\Rightarrow\quad \frac{x^2}{4} = 25 \quad\Rightarrow\quad x^2 = 100 \quad\Rightarrow\quad x = \pm 10$$

$x = -10$ makes $\log_5 x$ undefined, so reject it. Check $x = 10$: $2\log_5 10 - \log_5 4 = \log_5 \tfrac{100}{4} = \log_5 25 = 2$. ✓

8. (Challenge) Solve $(\log x)^2 - 3\log x + 2 = 0$.

Solution

This is a quadratic in $\log x$. Let $u = \log x$:

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

$\log x = 1$ gives $x = 10$, and $\log x = 2$ gives $x = 100$. Both are positive, so both work.

Check $x = 100$: $2^2 - 3(2) + 2 = 0$. ✓

9. (Challenge) Solve $\log_2 x + \log_4 x = 6$.

Solution

The bases are different, so first write $\log_4 x$ in base $2$. By change of base, $\log_4 x = \dfrac{\log_2 x}{\log_2 4} = \tfrac{1}{2}\log_2 x$. Then:

$$\log_2 x + \tfrac{1}{2}\log_2 x = 6 \quad\Rightarrow\quad \tfrac{3}{2}\log_2 x = 6 \quad\Rightarrow\quad \log_2 x = 4 \quad\Rightarrow\quad x = 16$$

Check: $\log_2 16 + \log_4 16 = 4 + 2 = 6$. ✓