Logarithmic Equations

Logarithmic equations are solved by respecting positive arguments, combining logs legally, and using their inverse exponential relationship.

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A logarithm reverses exponential growth. A slowly increasing curve lies to the right of a dashed vertical asymptote. The diagonal line y equals x is included as the reflection line used when comparing a logarithm with its exponential inverse.

Solve the algebra, then enforce the logarithm's domain: every argument must remain positive.

Domain conditions first

For real logarithms, each argument must satisfy

\text{argument}>0.

Record these restrictions before combining or exponentiating. Algebraic manipulation can produce candidates that make an original logarithm undefined.

Convert one logarithm to exponential form

\log_b A=c,

then

$$ A=b^c. $$

For example,

\log_3(x-1)=2

becomes $x-1=9$, so $x=10$. Since $x-1>0$, the solution is valid.

Use the one-to-one property

If two logs with the same valid base are equal,

\log_b A=\log_b B,

then $A=B$, provided $A>0$ and $B>0$.

Thus $\log_5(2x-1)=\log_5(x+7)$ gives $2x-1=x+7$, so $x=8$, which satisfies both argument conditions.

Worked example with an extraneous root

The rejected value solves the transformed quadratic but not the original logarithmic equation.

Common mistakes

Ignoring argument restrictions. Check every original logarithm.

Splitting $\log(M+N)$. No sum law exists.

Dropping coefficients instead of turning them into powers. Use the power law.

Keeping every polynomial root. Transformed equations can introduce invalid candidates.

Rounding before verification. Preserve precision, then check the original.

Quick self-check

What inequalities keep all original arguments positive?
Can the logs be combined legally?
Should I convert to exponential form or use one-to-one reasoning?
Have all algebraic candidates been tested against the domain?
If the answer is approximate, does it verify to the requested precision?