Question

How do you evaluate `log_6(1/36)`?

Answer 1

`log_6(1/36)= -2`

`color(brown)("Another way of dealing with this type of problem!")`

Explanation:

Let some unknown value be `x`

Set `log_6(1/36) = x`.......(1)

Then `log_6(1/36) =x -> 6^x=1/36`.....(2)

We know that `6^2 = 6xx6 = 36`

So `1/(6^2) =1/6xx1/6=1/36`

Also another way of writing `1/6^2" is " 6^(-2) => x=-2`..... (3)

Substitute (3) into (2) giving:

From this we have: `6^(-2)=1/36`

Thus: `log_6(1/36)= -2`