Question 838674
Let {{{u=3^x}}}, {{{3^(2+x)=9*3^x=9u}}}, {{{3^(2-x)=9/3^x=9/u}}}
{{{3^(2+x) + 3^(2-x) = 82}}}
{{{9u+9/u=82}}}
{{{9u^2+9=82u}}}
{{{9u^2-82u+9=0}}}
{{{(9u-1)(u-9)=0}}}
Two solutions:
{{{9u-1=0}}}
{{{9u=1}}}
{{{u=1/9}}}
{{{3^x=1/9}}}
{{{x=-2}}}
and
{{{u-9=0}}}
{{{u=9}}}
{{{3^x=9}}}
{{{x=2}}}