Question 1098171
<br>Combining the two pieces of given information, we have<br>
{{{5^(4x) + 20*5^(2x) - 125) = 0}}}<br>
Since {{{(5^(2x))^2 = 5^(4x)}}}, this is simply a "quadratic" equation, where the "variable" is {{{5^(2x)}}}.  Factoring,<br>
{{{(5^(2x)+25)*(5^(2x)-5) = 0}}}
{{{5^(2x) = -25}}} or {{{5^(2x) = 5}}}<br>
The first of these gives no solution for x; the second one gives us
{{{5^(2x) = 5 = 5^1}}}
{{{2x = 1}}}
{{{x = 1/2}}}<br>
Solution: x = 1/2