Question 1195397
<br>
{{{5^(2x+3)-6(5^x )=595}}}<br>
The equation involves 5 to the powers (2x+3) and x.  Rewrite the equation as a "quadratic" with "variable" 5^x by writing {{{5^(2x+3)}}} as {{{(5^3)(5^(2x))}}}.<br>
{{{125(5^(2x))-6(5^x)-595=0}}}<br>
The quadratic does not factor over the integers; a graphing calculator shows<br>
{{{5^x=2.2058744}}} to several decimal places.  Then<br>
{{{x*log(5)=log(2.2058744)}}}
{{{x=log(2.2058744)/log(5)=0.4915529686}}} again to several decimal places.<br>
ANSWER: x = approximately 0.4915529686<br>