Question

Log 6 (1/36) 6 is the base. How to evaluate this?

Answer 1

We can write the argument of the logarithm as a power of 6:

$\log_6\dfrac1{36}=\log_6\dfrac1{6^2}=\log_66^{-2}$

Then using the property that $\log_ba^n=n\log_ba$, we get

$\log_6\dfrac1{36}=-2\log_66$

and since $6=6^1$, we have $\log_66=1$, so the value of this expression is simply -2.