Answer
$1.442$
Work Step by Step
Using change-of-base, $\displaystyle \log_{b}M=\frac{\log_{a}M}{\log_{a}b}$, we change the base to 10
$\displaystyle \log_{12}36=\frac{\log 36}{\log 12}$
... next, we recognize $36=6^{2}=2^{2}\cdot 3^{2}$
... so $\log 36 = 2\log 2+2\log 3$
.... also, $12=3\cdot 4=2^{2}\cdot 3$
... so $\log 12=2\log 2+\log 3$
... we applied $\log_{a}M^{p}=p\cdot\log_{a}M$ and $\log_{a}(MN)=\log_{a}M+\log_{a}N$
$... =\displaystyle \frac{2\log 2+2\log 3}{2\log 2+\log 3}$
... substitute $\log 2=0.301$ and $\log 3=0.477,$
$=\displaystyle \frac{2\times 0.301+2\times 0.477}{2\times 0.301+ 0.477}\approx$1.442075996
To three decimal places,
$=1.442$
