Solving Logarithmic Equations and Checking Validity

What it means to “solve” a logarithmic equation

Solving a logarithmic equation means finding all real values of the variable that make the equation true and keep every logarithm defined. Every log expression log_b(A) requires A > 0 (and the base conditions b > 0, b ≠ 1 are assumed fixed). Because algebraic steps can introduce values that make a log argument nonpositive, you must always check candidate solutions in the original equation.

1) Single-log equations: convert to exponential form

If the equation has one logarithm and it equals a number (or a simple expression), the fastest route is to rewrite in exponential form.

Example 1: basic conversion

Solve log_3(x) = 4.

• Convert to exponential form: x = 3^4.
• Compute: x = 81.
• Domain check: the argument is x, so require x > 0. Since 81 > 0, it is valid.

Example 2: single log with a linear argument

Solve log_5(2x - 1) = 3.

• Domain restriction first: 2x - 1 > 0 so x > 1/2.
• Convert: 2x - 1 = 5^3.
• Solve: 2x - 1 = 125 ⇒ 2x = 126 ⇒ x = 63.
• Check domain: 63 > 1/2, valid.

Example 3: log equals an expression (still single-log)

Solve log_2(x - 3) = log_2(5) + 1.

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• Domain: x - 3 > 0 ⇒ x > 3.
• Rewrite the right side as a single number: 1 = log_2(2), so log_2(5) + 1 = log_2(5) + log_2(2) = log_2(10).
• Now log_2(x - 3) = log_2(10) ⇒ x - 3 = 10 ⇒ x = 13.
• Check: 13 > 3, valid.

2) Multi-log equations: combine logs, then convert

When multiple logs appear on one side, use log properties to combine them into a single logarithm, then convert to exponential form (or equate arguments if logs with the same base are equal).

Example 4: sum of logs

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

• Domain: x > 0 and x - 2 > 0 ⇒ x > 2.
• Combine: log_2(x) + log_2(x - 2) = log_2(x(x - 2)).
• Convert: log_2(x(x - 2)) = 3 ⇒ x(x - 2) = 2^3 = 8.
• Solve: x^2 - 2x - 8 = 0 ⇒ (x - 4)(x + 2) = 0 ⇒ x = 4 or x = -2.
• Domain check: must have x > 2, so x = 4 is valid and x = -2 is rejected.

Example 5: difference of logs

Solve log_3(x + 1) - log_3(x - 2) = 2.

• Domain: x + 1 > 0 and x - 2 > 0 ⇒ x > 2.
• Combine: log_3((x + 1)/(x - 2)) = 2.
• Convert: (x + 1)/(x - 2) = 3^2 = 9.
• Solve: x + 1 = 9x - 18 ⇒ 19 = 8x ⇒ x = 19/8.
• Check: 19/8 = 2.375 > 2, valid.

Example 6: coefficient in front of a log

Solve 2log_5(x) = 1.

• Domain: x > 0.
• Use the power rule: 2log_5(x) = log_5(x^2).
• Convert: log_5(x^2) = 1 ⇒ x^2 = 5^1 = 5.
• Candidate solutions: x = ±√5.
• Domain check: x > 0, so x = √5 only.

3) Logs on both sides and different bases

When logs appear on both sides, first try to rewrite so the bases match. If bases differ, use change-of-base to compare them in a common base (often natural log or base 10), or rewrite one side using exponent relationships.

Example 7: same base on both sides

Solve log_7(2x - 3) = log_7(x + 5).

• Domain: 2x - 3 > 0 ⇒ x > 3/2; and x + 5 > 0 ⇒ x > -5. Combined: x > 3/2.
• Same base and both logs defined, so equate arguments: 2x - 3 = x + 5.
• Solve: x = 8.
• Check domain: 8 > 3/2, valid.

Example 8: different bases by rewriting

Solve log_2(x) = log_4(8).

• Evaluate the right side by rewriting base 4 as 2^2: log_4(8) = log_{2^2}(2^3) = 3/2.
• So log_2(x) = 3/2 ⇒ x = 2^{3/2} = √(2^3) = 2√2.
• Domain check: x > 0, valid.

Example 9: change-of-base to solve a variable-in-log equation

Solve log_3(x) = log_5(25).

• Compute the right side: log_5(25) = 2.
• Then log_3(x) = 2 ⇒ x = 3^2 = 9.

Now a case where change-of-base is used structurally:

Solve log_2(x) = log_3(x - 1).

• Domain: x > 0 and x - 1 > 0 ⇒ x > 1.
• Change-of-base: log_2(x) = ln(x)/ln(2) and log_3(x - 1) = ln(x - 1)/ln(3).
• Set equal: ln(x)/ln(2) = ln(x - 1)/ln(3).
• Cross-multiply: ln(3)·ln(x) = ln(2)·ln(x - 1).
• This is not usually solvable by simple algebra; it’s a good place to use numerical methods. For example, define f(x) = ln(3)ln(x) - ln(2)ln(x - 1) on x > 1 and approximate a root. (In many courses, problems are chosen to avoid this situation unless numerical solving is allowed.)

4) Domain checks and extraneous solutions

Two common ways extraneous solutions appear:

• You solve an equation that implicitly assumed log arguments were positive, but your algebra produces a value that violates that.
• You exponentiate both sides after manipulating logs, producing an equation that has solutions outside the original domain.

Example 10: extraneous solution from domain violation

Solve log_2(x - 1) + log_2(x - 3) = 1.

• Domain: x - 1 > 0 and x - 3 > 0 ⇒ x > 3.
• Combine: log_2((x - 1)(x - 3)) = 1.
• Convert: (x - 1)(x - 3) = 2.
• Expand: x^2 - 4x + 3 = 2 ⇒ x^2 - 4x + 1 = 0.
• Solve: x = (4 ± √(16 - 4))/2 = (4 ± √12)/2 = 2 ± √3.
• Check domain x > 3: 2 + √3 ≈ 3.732 is valid; 2 - √3 ≈ 0.268 is rejected.

Example 11: checking by substitution (recommended habit)

Solve log_10(2x) = log_10(x - 4) + 1.

• Domain: 2x > 0 ⇒ x > 0; and x - 4 > 0 ⇒ x > 4. Combined: x > 4.
• Rewrite 1 as log_10(10): right side becomes log_10(x - 4) + log_10(10) = log_10(10(x - 4)).
• So log_10(2x) = log_10(10(x - 4)) ⇒ 2x = 10(x - 4).
• Solve: 2x = 10x - 40 ⇒ -8x = -40 ⇒ x = 5.
• Substitute into the original: left log_10(10) = 1; right log_10(1) + 1 = 0 + 1 = 1. Works, and 5 > 4.

5) Mixed exercises (increasing complexity, including parameters)

For each problem: (i) state the domain restrictions, (ii) solve, (iii) check candidates in the original equation.

A. Warm-up (single log)

• log_4(x) = -2
• log_7(3x + 1) = 0
• log_2(x - 5) = 3

B. Combine then convert (multi-log)

• log_3(x) + log_3(x - 6) = 2
• log_5(x + 4) - log_5(x) = 1
• 3log_2(x) - log_2(4) = 5

C. Logs on both sides

• log_6(2x - 1) = log_6(5x - 10)
• log_2(x) = log_8(4x)
• log_9(x - 2) = log_3(5)

D. Designed to produce an extraneous candidate

• log_2(x - 2) + log_2(x - 8) = 4
• log_3(x + 2) = log_3(2x - 1) (be careful: domain can eliminate a solution even if the linear equation looks fine)

E. Parameterized examples

These train you to keep the domain in view while solving symbolically.

E1: a parameter inside the argument

Solve for x in terms of a: log_2(x - a) = 3.

• Domain: x - a > 0 ⇒ x > a.
• Convert: x - a = 8 ⇒ x = a + 8.
• Check: a + 8 > a always, so the solution is valid for all real a.

E2: parameter affects whether a solution exists

Solve for x in terms of k: log_3(x) + log_3(x - k) = 2.

• Domain: x > 0 and x - k > 0 ⇒ x > max(0, k).
• Combine: log_3(x(x - k)) = 2 ⇒ x(x - k) = 9.
• Quadratic: x^2 - kx - 9 = 0 ⇒ x = (k ± √(k^2 + 36))/2.
• Validity: since √(k^2 + 36) > |k|, the “minus” root (k - √(k^2 + 36))/2 is always negative, so it fails x > 0. The “plus” root is positive and must also satisfy x > k when k > 0; it does, because (k + √(k^2 + 36))/2 > k.
• So the valid solution is x = (k + √(k^2 + 36))/2 for all real k.

E3: parameter in the exponent after converting

Solve for x: log_b(x) = m where b > 0, b ≠ 1.

• Domain: x > 0.
• Convert: x = b^m.
• Check: b^m > 0 always, so it is valid.