Answer
$2.452$
Work Step by Step
Using change-of-base, $\displaystyle \log_{b}M=\frac{\log_{a}M}{\log_{a}b}$, we change base to 10
$\displaystyle \log_{6}81=\frac{\log 81}{\log 6}$
... next, we recognize $81=3^{4}$, and apply $\log_{a}M^{p}=p\cdot\log_{a}M$.
... Also, we write $6=2\cdot 3$ and apply $\log_{a}(MN)=\log_{a}M+\log_{a}N$
$... =\displaystyle \frac{4\log 3}{\log 2+\log 3}$
... substitute $\log 2=0.301$ and $\log 3=0.477,$
$=\displaystyle \frac{4(0.477)}{0.301+0.477}\approx 2.45244215938$
To three decimal places,
$=2.452$
