Solving Logarithmic Equations

Algebra 2 · Axiom Academy

LESSON Solving Logarithmic Equations From to b^y = x — condense the logs, convert to exponential form, then reject any root that makes an argument negative. 1. Convert to Exponential Form The definition of a logarithm is a two-way street. A log is an exponent, so any statement about a log can be rewritten as a statement about a power — and the power is what you can solve. Watch the value of the log swing up to become the exponent. The log's value is the exponent on the base Try it on : the base is 2 , the value is 5 , so x must be 2 raised to that value. Since x = 32 > 0 , the single argument is positive, so x = 32 is genuine. When several logs appear, you cannot convert yet — there must be one log to peel off. The log properties let you fuse them: a sum of logs becomes the log of a product. Watch two logs merge into one. Condense with the product rule, then convert and solve. Both x = 3 and x = -1 are candidates — the algebra is not the final word. We must still test them against the domain (Step 4). 3. Equal Logs, Equal Arguments Sometimes both sides are already a single log with the same base. Because a logarithm is one-to-one — each output comes from exactly one input — equal logs force their arguments to be equal. The matching log shells simply lift away. Solve by matching the arguments. Check: at x = 4 both arguments equal 7 > 0 , so x = 4 is valid. 4. Check for Extraneous Solutions

This is the written version of the interactive lesson above. See the full Algebra 2 course.