Question 54932
This might not be the shortest solution, but it seems to check OK
{{{3^x+5=9^x}}}
note that {{{9^x = (3^2)^x}}}, so
{{{3^x + 5 = (3^2)^x}}}
also, {{{(3^2)^x = 3^(2x)}}} because, if x were 5, let's say, then
{{{(3^2)^5 = (3^2)*(3^2)*(3^2)*(3^2)*(3^2) = 3^(2+2+2+2+2) = 3^(2*5)}}}
{{{3^x + 5 = 3^(2x)}}}
subtract 3^x from both sides
{{{3^(2x) - 3^x = 5}}}
note that {{{3^(2x) = 3^(x+x) = (3^x)*(3^x)}}}
{{{(3^x)*(3^x) - 3^x = 5}}}
now factor out 3^x
{{{(3^x)*(3^x - 1) = 5}}}
make a substitution, let {{{z = 3^x}}}, so
{{{z(z - 1) = 5}}}
{{{z^2 - z = 5}}}
solve completing the square
{{{z^2 - z + 1/4 = 5 + 1/4}}}
{{{z - 1/2)^2 = 21/4}}}
take the square root of both sides
{{{z - 1/2 = sqrt(0 + 21)/2}}}
{{{z = (1 + sqrt(0 + 21)) / 2}}}
{{{z = 5.582576 / 2}}}
{{{z = 2.7912878}}}
since {{{z = 3^x}}}
{{{3^x = 2.7912878}}}
take the log of both sides. I chose log to the base e.
{{{x*ln(3) = 1.0265031}}}
{{{1.0986123*x = 1.0265031}}}
{{{x = 1.0265031 / 1.0986123}}}
{{{x = .9343634}}} answer
I'll sub this back into the original equation
{{{3^x+5=9^x}}}
{{{3^.9343634 = 2.7912879}}}
{{{9^.9343634 = 7.7912883}}}
{{{ 2.7912879 + 5 = 7.7912883}}}
This is true up to the 5th decimal place, so I believe the error is due
to my rounding off on calculator
  Note also that I could have used the negative square root of 21
in the calculation for z, but this would have made {{{3^x}}} negative,
and that's impossible, no matter what x is.