Question 630598
<pre>
{{{(1/2)^(2x-1)=(1/4)^(3x+2)}}}

Write {{{(1/4)}}} as {{{(1/2)^2}}}

{{{(1/2)^(2x-1)=((1/2)^2)^(3x+2)}}}

Remove outer parentheses on right by multiplying exponents:

{{{(1/2)^(2x-1)=(1/2)^(6x+4)}}}

Since the base {{{1/2}}} is the same on both sides and since {{{1/2}}} is
positive and not equal to 1, we may equate exponents:

2x-1 = 6x+4
 -4x = 5
   x = {{{-5/4}}}

No need for logs on this problems.  Logs could be used,
but they are unnecessary.

Edwin</pre>