Quote:
Originally Posted by Dragonballful23
When is the following true?
3^x+1=9^x

Did you mean 3^(x+1) = 9^x?
If so, then:
3^(x+1) = (3^2)^x
3^(x+1) = 3^(2x)
x+1 = 2x
x = 1
But the way the problem is originally written is much harder to solve.
(3^x) + 1 = 9^x
(3^x) + 1 = 3^(2x)
1 = 3^(2x)  3^x
3^(2x)  3^x  1 = 0
Let a = 3^x
a^2  a  1 = 0
Then using the quadratic formula a = (1 + sqrt(5))/2 which is the golden ratio!
So 3^x = (1 + sqrt(5))/2
Taking log base 3 (log3) of both sides gives:
x = log3[(1 + sqrt(5))/2]
or using natural logs x = [ln(1 + sqrt(5))  ln(2)]/(ln(3))