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\frac1{36}=\log_6\frac1{6^2}=\log_66^(-2)`

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

`\log_6\frac1{36}=-2\log_66`

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