Question 978393
<pre>
{{{32^x}}}{{{""=""}}}{{{8^(2x-1)}}}

No logarithms are needed, although you can get the answer
with logs.  But I won't use logs here:

Write 32 as 2<sup>5</sup> and 8 as 2<sup>3</sup>

{{{(2^5)^x}}}{{{""=""}}}{{{(2^3)^(2x-1)}}}

Remove the parentheses by multiplying the exponents:

{{{2^(5x)}}}{{{""=""}}}{{{2^(3(2x-1))}}}

The bases are equal, non-negative and not equal to 1, so
the exponents are equal, so we can drop the bases:

{{{5x}}}{{{""=""}}}{{{3(2x-1)}}}

You finish.

Edwin</pre>