Solving Logarithmic Equations

Core Idea

Logarithmic equations contain logarithmic expressions with the variable in the argument. Two main strategies: (1) If the equation has a single log on each side with the same base, set the arguments equal: log_b(A) = log_b(B) implies A = B. (2) If the equation has log = constant, convert to exponential form: log_b(A) = c means A = b^c. Use log properties to condense multiple log terms first. Always check for extraneous solutions (arguments of log must be positive).

How It's Best Learned

Practice converting between log and exponential forms. Solve equations by condensing log expressions using properties, then converting. Emphasize domain checking: solutions must make all original log arguments positive. Give examples where extraneous solutions arise.

Common Misconceptions

• Not checking domain restrictions (log of a negative number is undefined).
• Forgetting to condense multiple log terms before converting to exponential form.
• Confusing log_b(x) = y with b*x = y (should be b^y = x).
• Thinking that because algebraic steps produce a valid number, it must be a valid solution (must check domain).

Explainer

Solving logarithmic equations rests on two ideas you already know: the definition of a logarithm as the inverse of exponentiation, and the log properties that let you combine or split logarithmic expressions. The definition says log_b(A) = c means exactly b^c = A — the log asks "what exponent on b gives A?" That relationship is the key to unlocking equations where the unknown is inside a logarithm.

There are two main situations. In the first, you have a single log equal to a number: log_b(expression) = c. Here you simply convert to exponential form — b^c = expression — and solve the resulting algebraic equation. For example, log₂(x + 3) = 4 becomes 2⁴ = x + 3, so x = 13. In the second situation, you have logs on both sides with the same base: log_b(A) = log_b(B). Since the log function is one-to-one (each output corresponds to exactly one input), equal outputs require equal inputs: A = B. So you set the arguments equal and solve. For log₃(2x + 1) = log₃(x + 4), you get 2x + 1 = x + 4, giving x = 3.

When an equation has multiple log terms, your first move should always be to condense them into a single log using log properties. The key properties are: log_b(MN) = log_b(M) + log_b(N), log_b(M/N) = log_b(M) − log_b(N), and log_b(M^r) = r·log_b(M). For example, to solve log₂(x) + log₂(x − 2) = 3, first condense the left side: log₂(x(x−2)) = 3, then convert: 2³ = x(x−2), giving 8 = x² − 2x, or x² − 2x − 8 = 0, so (x−4)(x+2) = 0, producing x = 4 and x = −2.

Here is where the domain check becomes critical. The argument of any logarithm must be strictly positive — log of zero or a negative number is undefined. Plug both candidates back into the original equation: if x = 4, the arguments x = 4 and x−2 = 2 are both positive, so x = 4 is valid. If x = −2, the argument x = −2 is negative, so x = −2 is extraneous — rejected. The algebra produced it, but it is not a solution to the original equation. Always check every candidate answer against the domain conditions in the original problem; extraneous solutions arise precisely because squaring or combining logs can obscure domain restrictions.