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Worked solution

Change both logarithms to the same base

Solve the equation log⁡3x+log⁡9x=6\log_3x+\log_9x=6.

CJC · O Level revision · Example 10

Answer

x=81x=81.

Compare your reasoning

Apply change of base

log⁡3x+log⁡3xlog⁡39=6\log_3x+\frac{\log_3x}{\log_39}=6

As the printed callout instructs, express log base 9 in terms of log base 3. Require x>0.

Evaluate the base conversion denominator

log⁡3x+log⁡3xlog⁡332=6 ⟹ log⁡3x+log⁡3x2log⁡33=6 ⟹ log⁡3x+log⁡3x2(1)=6\begin{aligned}&\log_3x+\frac{\log_3x}{\log_33^2}\\ & =6\ \Longrightarrow\ \log_3x+\frac{\log_3x}{2\log_33}\\ & =6\ \Longrightarrow\ \log_3x+\frac{\log_3x}{2(1)}\\ & =6\end{aligned}

The source expands 9=3², then uses log₃3=1.

Solve and return to index form

32log⁡3x=6 ⟹ log⁡3x=4 ⟹ x=34=81\frac32\log_3x=6\ \Longrightarrow\ \log_3x=4\ \Longrightarrow\ x=3^4=81

The positive result satisfies both logarithms.

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