Question 984083
<pre>
{{{log((2))/log((32))}}}

Since 32 = (2)(2)(2)(2)(2) = 2<sup>5</sup>

{{{log((2))/log((2^5))}}}

A rule of lagarithms tells us that we can move the exponent 5
out front as a coefficient of the log:

{{{log((2))/(5log((2)))}}}

Now we can cancel the log 2's and get

{{{cross(log((2)))/(5cross(log((2))))}}}

which is just

{{{1/5}}}

Now do yours the exact same way.

Edwin</pre>