Logarithm problem: 6^x + 4^x = 9^x SOLVE FOR X?

Answer 1

6^x + 4^x = 9^x

First, let's split up each base as its prime factors.

(3 * 2)^x + (2 * 2)^x = (3 * 3)^x

Now, since (a*b)^m = (a^m)(b^m),

(3^x)(2^x) + (2^x)(2^x) = (3^x)(3^x)

Move everything over to the left hand side, getting

(2^x)(2^x) + (3^x)(2^x) - (3^x)(3^x) = 0

I thought I knew where to go from here, but I don't; sorry.

Answer 2

6^x + 4^x = 9^x
<=>(2*3)^x + 2^(2x) = 3^(2x)
<=> (2/3)^x + (2/3)^(2x) = 1 (divide by 3^(2x))
let t be (2/3)^x, t > 0
=> t^2 + t - 1 =0
=> t = (-1 + sqrt(5)) / 2 or t = (-1 + sqrt(5)) / 2
=> (2/3)^x = (-1 + sqrt(5))/2
=> x = log base 2/3 of (-1 + sqrt(5))/2