Question 1098778
<br>With the "5^2x" and "5^x" terms in the equation, this can be viewed as a "quadratic" equation in which the "variable" is 5^x, because {{{(5^x)^2 = 5^(2x)}}}.<br>
So rewrite the equation as
{{{5^(2x) + 5^x - 3 = 0}}}<br>
If it helps ease the confusion, you can define a new variable {{{u = 5^x}}} and write the equation as {{{u^2 + u - 3 = 0}}}<br>
The quadratic does not factor, so use the quadratic formula:
{{{5^x = (-1+-sqrt(13))/2}}}<br>
Since 5^x is never negative, we choose the positive solution: {{{5^x = (-1+sqrt(13))/2}}}<br>
Then to solve for x, since it is an exponent, we take logs of both sides:
{{{x*log(5) = log((-1+sqrt(13))/2)}}}
{{{x = log((-1+sqrt(13))/2)/log(5)}}}