How to Solve Logarithmic Equations (Step by Step)

A logarithm answers the question "what exponent do I need?"

$$\log_b x = y \quad \Longleftrightarrow \quad b^y = x$$

For example, $\log_2 8 = 3$ because $2^3 = 8$.

Two notation reminders:

• $\log x$ with no base means base 10.
• $\ln x$ means base $e$, so $\ln x = y$ means $e^y = x$.

The domain rule: you can only take the log of a positive number. The input of every log in the equation (called the argument) must be greater than 0. This is why you must check your answers.

For a positive base $b \neq 1$ and positive $M$ and $N$:

• Product rule: $\log_b(MN) = \log_b M + \log_b N$
• Quotient rule: $\log_b\left(\frac{M}{N}\right) = \log_b M - \log_b N$
• Power rule: $\log_b(M^k) = k\log_b M$
• Special values: $\log_b b = 1$ and $\log_b 1 = 0$
• Change of base: $\log_b M = \frac{\ln M}{\ln b}$

There is no rule for $\log_b(M + N)$. It is not $\log_b M + \log_b N$.

Rewrite the equation in exponential form and solve.

Example 1: Solve $\log_2(x + 3) = 5$.

1. Rewrite: the base is 2, the exponent is 5, so $2^5 = x + 3$.
2. Simplify: $32 = x + 3$, so $x = 29$.
3. Check: $29 + 3 = 32 > 0$, and $\log_2 32 = 5$. It works.

Example 2: Solve $\ln(2x) = 3$.

1. Rewrite: $e^3 = 2x$.
2. Solve: $x = \frac{e^3}{2} \approx 10.04$.
3. Check: $2x = e^3 > 0$. It works.

Use the log rules to combine everything into one log, then rewrite in exponential form.

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

1. Combine with the product rule: $\log_2\left[x(x - 2)\right] = 3$.
2. Rewrite in exponential form: $x(x - 2) = 2^3 = 8$.
3. Expand and set to 0: $x^2 - 2x - 8 = 0$.
4. Factor: $(x - 4)(x + 2) = 0$, so $x = 4$ or $x = -2$.
5. Check $x = 4$ in the original: $\log_2 4 + \log_2 2 = 2 + 1 = 3$. It works.
6. Check $x = -2$: $\log_2(-2)$ is undefined. Reject it.

Answer: $x = 4$. The value $x = -2$ is an extraneous solution: it solves the combined equation but not the original one.

If both sides are a single log with the same base, the arguments must be equal:

$$\log_b M = \log_b N \quad \Longrightarrow \quad M = N$$

Example 1: Solve $\log(3x - 1) = \log(x + 7)$.

1. Set the arguments equal: $3x - 1 = x + 7$.
2. Solve: $2x = 8$, so $x = 4$.
3. Check: $3(4) - 1 = 11 > 0$ and $4 + 7 = 11 > 0$. It works.

Example 2: Solve $\log_3(x^2 - 4) = \log_3(3x)$.

1. Set the arguments equal: $x^2 - 4 = 3x$.
2. Rearrange: $x^2 - 3x - 4 = 0$.
3. Factor: $(x - 4)(x + 1) = 0$, so $x = 4$ or $x = -1$.
4. Check $x = 4$: $16 - 4 = 12 > 0$ and $3(4) = 12 > 0$. It works.
5. Check $x = -1$: $(-1)^2 - 4 = -3$, which is negative. Reject it.

Answer: $x = 4$.

Example 3 (quotient rule): Solve $\ln(x + 6) - \ln x = \ln 4$.

1. Combine the left side: $\ln\left(\frac{x + 6}{x}\right) = \ln 4$.
2. Set the arguments equal: $\frac{x + 6}{x} = 4$.
3. Multiply by $x$: $x + 6 = 4x$, so $x = 2$.
4. Check: $\ln 8 - \ln 2 = \ln\frac{8}{2} = \ln 4$. It works.

Combining logs can create answers that do not work in the original equation, because $\log_b x + \log_b(x - 2)$ requires both $x > 0$ and $x - 2 > 0$, while $\log_b[x(x - 2)]$ only needs their product to be positive.

Checking routine:

1. Go back to the original equation, not your simplified one.
2. Plug each candidate into the argument of every log.
3. If any argument is zero or negative, reject that candidate.
4. Optionally, evaluate both sides to confirm they match.

Watch the power rule too. For $2\log_5 x = \log_5 16$, rewriting the left side as $\log_5(x^2)$ gives $x^2 = 16$, so $x = \pm 4$. But the original equation contains $\log_5 x$, so $x = -4$ is extraneous. The answer is $x = 4$.

Paste an equation into the VeilGPT solver to check your steps and your domain checks.