Solving Logarithmic Equations

You have log₃(x) = 4. The correct answer is x = 81, but many students write x = 64 because they treat "log" as a variable. A logarithm is an exponent: logᵢ(x) = y means b^y = x. That fact is the only tool you need for solving logarithmic equations. Every method returns to that definition. The most common mistake is thinking log(x) is something you can move around; it is not. You cannot add log(x) + 3 to get log(x+3), and you cannot divide log(x) by 2 to get log(x/2). Each step must be justified by a property of logarithms or by the exponential inverse.

Rewrite in Exponential Form

When your equation has a single logarithm, rewrite it in exponential form. The equation logᵢ(x) = y becomes x = b^y. No condensing, no combining, no moving terms, just the inverse property.

Example: log₂(3x + 1) = 5. Rewrite as 3x + 1 = 2⁵. Solve: 3x + 1 = 32, so 3x = 31, x = 31/3. That is the candidate. You still need to check it, see the section on extraneous solutions below.

This works for any base. log(x) = 2.3010 means x = 10^2.3010 ≈ 200. ln(x) = 1.5 means x = e^1.5 ≈ 4.48. The exponential form removes the log in one step.

Failure case: If you get log₂(, 5) in a check, the equation has no real solution. The argument of a log must be positive. Domain: x > 0.

Use the Product, Quotient, and Power Rules

When you have two or more logs on the same side, condense them into a single logarithm using the product, quotient, and power rules before rewriting as exponential.

Product rule: logᵢ(M) + logᵢ(N) = logᵢ(MN).
Quotient rule: logᵢ(M), logᵢ(N) = logᵢ(M/N).
Power rule: logᵢ(M^p) = p · logᵢ(M).

Example: log₄(x) + log₄(x, 3) = 1. Condense: log₄[x(x, 3)] = 1. Rewrite as exponential: x(x, 3) = 4¹ = 4. Solve: x², 3x = 4, so x², 3x, 4 = 0, (x, 4)(x + 1) = 0, giving x = 4 or x =, 1. Check: x =, 1 gives log₄(, 1) which is undefined. Only x = 4 works.

Do not try to separate a sum inside a log, log(a+b) does not equal log(a) + log(b). The product rule applies to multiplication, not addition.

Drop Logs Only When Bases Match

When your equation has a single log on each side, you can often drop the logs, but only if the bases are the same. If logᵢ(f(x)) = logᵢ(g(x)), then f(x) = g(x).

Example: log₃(2x + 1) = log₃(7). Drop the logs: 2x + 1 = 7, so 2x = 6, x = 3. Check: log₃(7) = log₃(7), works.

If the bases differ, use the change-of-base formula to rewrite them with a common base. For log₁₀(x) = log₂(5), convert both to natural logs: ln(x)/ln(10) = ln(5)/ln(2). Solve for x: ln(x) = ln(10) · [ln(5)/ln(2)], then x = e^{[ln(10) · ln(5)/ln(2)]}.

Failure case: log₂(x) = log₃(x). Dropping logs here would give x = x, which is true for all x, but the original equation is only true at x = 1. Check: log₂(1) = 0, log₃(1) = 0. Any other value fails.

Always Verify Against the Original Equation

Every solution you find by algebra must be checked against the original equation. The algebraic steps, squaring, combining logs, rewriting as exponential, can introduce solutions that are not valid for the original domain.

Domain rule: The argument of every logarithm must be greater than zero. If a candidate makes any argument zero or negative, it is extraneous.

Example: log₂(x, 2) + log₂(x, 1) = 1. Condense: log₂[(x, 2)(x, 1)] = 1. Rewrite as exponential: (x, 2)(x, 1) = 2. Solve: x², 3x + 2 = 2, so x², 3x = 0, x(x, 3) = 0, giving x = 0 or x = 3. Check: x = 0 gives log₂(, 2), undefined. x = 3 gives log₂(1) + log₂(2) = 0 + 1 = 1, works. The extraneous solution x = 0 came from the algebraic expansion, not from the original log equation.

Always substitute back into the original equation, not a simplified version. A student who checks against the condensed form would miss the domain error.

Example 1: Single Log, One Variable

Solve log₃(5x, 2) = 3. Rewrite as exponential: 5x, 2 = 3³ = 27. Solve: 5x = 29, x = 29/5 = 5.8. Check: log₃(5·5.8-2) = log₃(29-2) = log₃(27) = 3. Valid.

Example 2: Condensing with the Product Rule

Solve log₂(x) + log₂(x + 4) = 5. Condense: log₂[x(x + 4)] = 5. Exponential: x(x + 4) = 2⁵ = 32. Solve: x² + 4x, 32 = 0. Use quadratic formula: x = [, 4 ± sqrt(16 + 128)]/2 = [, 4 ± sqrt(144)]/2 = [, 4 ± 12]/2. So x = 4 or x =, 8. Check: x = 4 gives log₂(4) + log₂(8) = 2 + 3 = 5, works. x =, 8 gives log₂(, 8), undefined. Extraneous.

Example 3: Logs on Both Sides

Solve log₅(3x + 1) = log₅(2x + 9). Drop the logs (same base): 3x + 1 = 2x + 9. Solve: x = 8. Check: log₅(25) = log₅(25), valid.

Example 4: Natural Log with a Coefficient

Solve 2·ln(x) = ln(4) + ln(x, 2). Use power rule: ln(x²) = ln[4(x, 2)]. Drop the logs (both base e): x² = 4x, 8. Solve: x², 4x + 8 = 0. Discriminant: 16-32 =, 16. No real solution. The equation has no solution in the real numbers.

Confusing ln and log

ln(x) uses base e ≈ 2.71828; log(x) typically means base 10. In pH calculations, pH =, log₁₀([H⁺]), using ln gives the wrong value by a factor of about 2.3026.

Misapplying the Product Rule

log(a + b) is not log(a) + log(b). The rule applies to products: log(ab) = log(a) + log(b).

Forgetting the Characteristic

When using four-figure log tables, the characteristic (integer part) is determined by the number's order of magnitude. For numbers between 0 and 1, the characteristic is negative. For log(0.003), the characteristic is, 3, mantissa about 0.4771, so log(0.003) ≈, 2.5229.

Interpolation Error

Linear interpolation on a log table assumes the function is linear between entries; the error is small (under 0.001) but real. For precise work, use a calculator.

• Base Constraint: b > 0, b ≠ 1
• Argument Constraint: x > 0
• Inverse Property: b^(log_b(x)) = x
• Product Rule: log_b(M) + log_b(N) = log_b(MN)
• Quotient Rule: log_b(M), log_b(N) = log_b(M/N)

For most solving logarithmic equations problems, a calculator is faster and more accurate. Four-figure log tables give log₁₀ values to four decimal places for numbers 1.000 to 9.999; for numbers outside that range, you must determine the characteristic yourself. The two-step process: find the mantissa from the table for the significant digits, then append the characteristic based on the number's magnitude.

Example: Find log₁₀(200) using a four-figure table. 200 = 2 × 10². Log₁₀(2) from the table: 0.3010. Characteristic: 2. So log₁₀(200) = 2.3010. A calculator gives the same value.

For antilog: given log =, 3.45, write, 3.45 =, 4 + 0.55. The characteristic is, 4, the mantissa is 0.55. Find the antilog of 0.55 from the table (≈ 3.548), then append the characteristic: 3.548 × 10⁻⁴ = 0.0003548.

Use a calculator for any base other than 10 or e. The change-of-base formula, log_b(x) = ln(x)/ln(b), works in any calculator with a ln key.

Number	log₁₀ (Common Log)	ln (Natural Log)	log₂ (Binary Log)
2	0.3010	0.6931	1.0000
3	0.4771	1.0986	1.5850
5	0.6990	1.6094	2.3219
7	0.8451	1.9459	2.8074
10	1.0000	2.3026	3.3219
e	0.4343	1.0000	1.4427

The single most practical thing: before you write any answer, check the domain of every logarithm in the original equation. If a candidate makes any argument ≤ 0, discard it immediately. That one habit eliminates the most common failure in solving logarithmic equations.