Question 54932
3^x+5=9^x
Rewrite as:
3^(2x)-3^x-5=0
Let 3^x be "w"
Rewrite the equation in "w", as follows:
w^2-w-5=0
Use the quadratic formula to solve for "w", as follows:
w=[1+-sqrt(1+20)]/2
w=[1+-sqrt(21)]/2
Convert back to "x" notation.
3^x=[1+-sqrt(21)]/2
Take the log of both sides to get:
x(log 3)=[1+-sqrt(21)]/2
x=[1+-sqrt(21)/[2log(3)]
Cheers,
Stan H.