How to Solve Logarithm Equations Step by Step (Examples)

To solve a logarithm equation, use the same routine every time: condense the logarithms into one with the log rules, rewrite the equation in exponential form, solve for x, then check your answer in the original equation. The check is not optional — condensing can produce answers that look right but are actually invalid. The key relationship is simple: log_b(x) = y means exactly the same thing as bʸ = x. So log₂(3x − 4) = 3 is really just 2³ = 3x − 4 in disguise, and it solves to x = 4. This guide covers the three log rules you need, then five worked examples that all follow this routine.

The 3 Log Rules You Need

Learn these three identities first — every example below uses at least one of them.

1. Product rule: log_b(M · N) = log_b(M) + log_b(N). A log of a product splits into a sum of logs — e.g. log₂(8 · 4) = 3 + 2 = 5. Read backward, it condenses two logs into one, which is what Examples 2 and 3 need.

2. Quotient rule: log_b(M / N) = log_b(M) − log_b(N). A log of a quotient splits into a difference of logs — e.g. log(1000 / 10) = 3 − 1 = 2.

3. Power rule: log_b(Mᵏ) = k · log_b(M). An exponent inside the log moves out front as a multiplier — e.g. log(10⁴) = 4 · 1 = 4.

Finally, the one relationship that turns logarithms into plain algebra: log_b(x) = y means exactly the same thing as bʸ = x. Once an equation is in this exponential form, the logarithm is gone and you just solve normally.

That conversion is where most log equations are won or lost, so practise flipping between the two forms until it feels automatic. If the powers slow you down, the exponent calculator checks values like 2³ = 8 and 10³ = 1000 instantly.

How to Solve Logarithm Equations: 5 Worked Examples

Example 1: Convert to Exponential Form — log₂(3x − 4) = 3

A single logarithm equal to a number: convert it straight into exponential form.

1. Rewrite in exponential form. log₂(3x − 4) = 3 means 2³ = 3x − 4.
2. Simplify the power. 2³ = 8, so 8 = 3x − 4.
3. Solve the linear equation. Add 4: 12 = 3x, then divide by 3: x = 4.
4. Check the domain. The argument must be positive: 3(4) − 4 = 8 > 0 ✓.

Answer: x = 4. Sanity check: log₂(8) = 3 because 2³ = 8. ✓

Example 2: Condense With the Product Rule — log(x) + log(x − 3) = 1

With log meaning base 10, condense the two added logs into one first.

1. Condense using the product rule. log(x) + log(x − 3) = log(x(x − 3)), so log(x(x − 3)) = 1.
2. Convert to exponential form. log(x(x − 3)) = 1 means 10¹ = x(x − 3), so 10 = x(x − 3).
3. Rearrange into a quadratic. x² − 3x = 10 becomes x² − 3x − 10 = 0.
4. Factor and solve. (x − 5)(x + 2) = 0 gives candidates x = 5 and x = −2.
5. Check the domain. The original equation needs x > 0 AND x − 3 > 0, i.e. x > 3. Only x = 5 qualifies; reject x = −2.

Answer: x = 5. Verify: log(5) + log(2) = log(10) = 1. ✓

Notice where the real work happened: the quadratic step. If factoring x² − 3x − 10 = 0 is the part that slows you down, the quadratic equation solver shows the factoring steps for any quadratic you enter.

Example 3: The Extraneous Solution Trap — log₂(x) + log₂(x − 2) = 3

Two candidates will emerge — but only one survives.

1. Condense using the product rule. log₂(x) + log₂(x − 2) = log₂(x(x − 2)), so log₂(x(x − 2)) = 3.
2. Convert to exponential form. log₂(x(x − 2)) = 3 means 2³ = x(x − 2), so 8 = x(x − 2).
3. Rearrange. x² − 2x = 8 becomes x² − 2x − 8 = 0.
4. Factor and solve. (x − 4)(x + 2) = 0 gives candidates x = 4 and x = −2.
5. Test each candidate in the ORIGINAL equation. For x = 4: log₂(4) + log₂(2) = 2 + 1 = 3 ✓, both arguments positive. For x = −2: you would need log₂(−2) + log₂(−4) — logs of negatives, which do not exist. Reject x = −2.

Answer: x = 4. Here is why the extraneous solution appeared: condensing widened the domain. The original expression needs both arguments positive, but the condensed log₂(x(x − 2)) only needs the product positive — true even when both factors are negative, which is how x = −2 sneaked in. Steps after condensing produce candidates, not confirmed solutions, so plugging back into the original equation is mandatory.

Example 4: Change of Base — Solve 2 · log(x) = 6

Handling a coefficient in front of the log, plus the change-of-base formula for verifying on any calculator.

1. Isolate the logarithm. Divide both sides by 2: log(x) = 3.
2. Convert to exponential form. log(x) = 3 means 10³ = x.
3. Evaluate. 10³ = 1000, so x = 1000.
4. Check the domain. The argument x = 1000 is positive — valid.
5. Verify with change of base. log_b(x) = log(x) / log(b) lets you evaluate any log with your calculator's log or ln buttons: ln(1000) ÷ ln(10) = 3 ✓. It also handles awkward bases, e.g. log₅(25) = ln(25) ÷ ln(5) = 2.

Answer: x = 1000. Check: 2 · log(1000) = 2 · 3 = 6 ✓.

Example 5: Logs on Both Sides — log(2x + 1) = log(x + 7)

Same base on both sides means the arguments must be equal.

1. Equate the arguments. 2x + 1 = x + 7.
2. Solve. Subtract x: x + 1 = 7, then subtract 1: x = 6.
3. Check both arguments. Both must be positive in the original equation: 2(6) + 1 = 13 > 0 and 6 + 7 = 13 > 0 ✓.

Answer: x = 6. Verify: log(13) = log(13) ✓. The shortcut only works when the logs are fully isolated, one per side — the reason the earlier examples needed condensing first.

Checking Your Answers: The Step That Catches Fake Solutions

1. Plug the candidate back into the original equation (not the condensed or converted version) and confirm both sides are equal.
2. Confirm every log argument is positive in the original equation. Any negative or zero argument disqualifies the candidate instantly.
3. Cross-check numerically with the free logarithm calculator — enter the original expression with your x value and confirm each log evaluates cleanly.

Common Mistakes to Avoid

• Dividing logs like ordinary numbers. log(a) / log(b) does NOT equal log(a / b) — that ratio is a change-of-base expression: log(a) / log(b) = log_b(a). The quotient rule only applies to the log of a quotient: log(a / b) = log(a) − log(b).
• Forgetting the domain. Skipping the domain check and accepting an extraneous solution is the most common error. Always require: every argument > 0, every base > 0 and ≠ 1.
• Misapplying the power rule to sums. log(x + y) is NOT log(x) + log(y). The rules apply to products, quotients, and powers — never to sums or differences inside the log.
• Dropping the base during conversion. log_b(x) = y becomes bʸ = x, not yᵇ. Remember: "base to the power of the answer equals the argument."
• Splitting logs over addition. log(3x) equals log(3) + log(x), not x · log(3) — the log splits the product; it never multiplies anything.

Frequently asked questions

What is the first step in solving a log equation?

Turn the log into something you can "see through." A single isolated log converts directly to exponential form: log_b(x) = y becomes bʸ = x. Multiple logs get condensed into one first (product, quotient, or power rule), then converted. Note the domain (every argument > 0) up front so you reject invalid answers later.

Why do extraneous solutions appear when solving logarithm equations?

Because condensing widens the domain. In log₂(x) + log₂(x − 2) = 3 both arguments must be positive, but condensed as log₂(x(x − 2)) = 3 only the product must be positive — also true when both factors are negative. Test every candidate in the original equation; reject any that make an argument zero or negative.

Can the argument of a logarithm be negative?

No. For real-number logarithms the argument must be strictly positive, and the base positive and not equal to 1. That is why domain checks matter: x = −2 can satisfy the converted quadratic yet still be invalid, because it would force you to evaluate log₂(−2).

What is the change-of-base formula, and when do I use it?

log_b(x) = log(x) / log(b), with the right-hand logs in any convenient base — usually base 10 or e. Use it to evaluate or verify logs with bases your calculator lacks, e.g. log₅(25) = ln(25) ÷ ln(5) = 2.

How can I check my answer quickly?

Substitute x into the original equation, confirm both sides match and every argument is positive. For a numeric double-check, enter the expression into the free logarithm calculator — it evaluates logs in any base instantly, so verifying x = 1000 in 2 · log(x) = 6 takes seconds.

Solving logarithm equations comes down to a repeatable habit: apply the log rules, convert to exponential form, solve the algebra, and always check the domain. Work through the five examples with pencil and paper, then change one number in each and solve again. To verify answers fast — or to handle unusual bases — use the free logarithm calculator. And if the plain algebra at the end of each example is what trips you up, our guide on how to solve linear equations covers it step by step.