Question 72954
3^x+5 = 9^x
9^x-3^x-5=0
3^(2x)-3^x-5=0
Let w = 3^x; then substitute to get:
w^2-w-5=0
Use the quadratic formula to solve for w:
w=[1+-sqrt(1-4*-5)]/2
w=[1+sqrt21]/2 or w = [1-sqrt21]/2
w=3.29129  or w=-0.645644

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Now substitute 3^x in place of w
3^x = 3.29129 or 3^x=-0.645644
But 3^x cannot be negative so we need only solve the 1st for "x":
Take the log to get:
xlog3 = log3.29129
x=log3.29129/log3
x=1.0843...
Cheers,
Stan H.