Question 1080363
{{{9^x=(3^2)^x=(3^x)^2}}}
So then substituting {{{u=3^x}}},
{{{u+u^2=1/27}}}
{{{u^2+u+1/4=1/27+1/4}}}
{{{(u+1/2)^2=31/108}}}
{{{u+1/2=0 +- sqrt(31/108)}}}
{{{u=-1/2 +- sqrt(31/108)}}}
{{{3^x=-1/2 +- sqrt(31/108)}}}
Only positive right hand sides will give a real answer,
{{{3^x=-1/2+sqrt(31/108)}}}
{{{x=log(3,(sqrt(93)/18-1/2)))}}}

{{{highlight(x=log((sqrt(93)/18-1/2))/log((3))))}}}