Question 1031012
<pre>
{{{(5^x)(5^x-1)=10}}}

{{{5^(2x)-5^x=10}}}

{{{5^(2x)-5^x-10=0}}}

Let {{{5^x=u}}}
Then {{{5^(2x)=u^2}}}

Substitute {{{u^2}}} for {{{5^(2u)}}} and {{{u}}} for {{{5^x}}}

 
{{{u^2-u-10=0}}}

{{{u = (-b +- sqrt(b^2-4ac))/(2a) }}} 

{{{u = (-(-1) +- sqrt( (-1)^2-4*(1)*(-10) ))/(2*1) }}} 

{{{u = (1 +- sqrt(1+40 ))/2 }}}

{{{u = (1 +- sqrt(41))/2 }}}

Substitute {{{5^x}}} for {{{u}}}

{{{5^x = (1 +- sqrt(41))/2}}}

Ignore the minus sign because 5<sup>x</sup> is positive:

Take the natural logarithm of both sides

{{{ln(5^x) = ln((1 + sqrt(41))/2)}}}

{{{x*ln(5) = ln((1 + sqrt(41))/2)}}}
 
{{{x = ln((1 +- sqrt(41))/2)/ln(5)}}}

{{{x = 0.813175155554523}}}, approximately.

Edwin</pre>