How do you evaluate $log_36 (1/6)$ ?

Question

4974 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-26T16:11:34

$log_36(1/6) = -1/2$

Recall that $1/6$ can be written as $6^-1$ .

$=log_36(6^-1)$

Now, use the change of base rule $log_a(n) = logn/loga$ .

$=log(6^-1)/log36$

$=log(6^-1)/log(6^2)$

Use the rule $loga^n = nloga$ .

$=(-1log6)/(2log6)$

$=-1/2$

Hopefully this helps!

How do you evaluate log_36 (1/6)log_36 (1/6)?