Question 744700
 9^(x)- 5(3^x) +6 =0  (hint let y = 3^x)

{{{(3^2)^x-5(3^x)+6=0}}}
{{{(3^x)^2-5(3^x)+6=0}}}, There is a rule of exponents for that.
Your hinted substitution:
{{{y^2-5y+6=0}}}, which is factorable into....
{{{(y-2)(y-3)=0}}}


Either y=2  OR   y=3
Reversing the substitution done,
Either {{{3^x=2}}}   Or   {{{3^x=3}}}


Whatever base you want, find logarithms of both sides for each choice of solution.

{{{log(3^x)=log(2)}}}
{{{x*log(3)=log(2)}}}
{{{highlight(x=log(2)-log(3))}}}, I skipped a step.  You might want that as {{{highlight(x=(log(2))/(log(3)))}}}


Do similarly for the other choice.