To solve this equation, we can rewrite it in terms of powers of 2, 3, and 6:

32^(2x) + 23^(2x) = 5(23)^x
32^(2x) + 23^(2x) = 52^x3^x
32^(2x) + 23^(2x) = 52^(x+1)3^(x)

Now, let y = 2^x and z = 3^x.

The equation becomes:
3y^2 + 2z^2 = 5yz
3y^2 + 2z^2 - 5yz = 0

(3y - z)(y - 2z) = 0

This gives us two possible solutions:

3y - z = 0y - 2z = 0

Now, we substitute back y = 2^x and z = 3^x to find the values of x:

3(2^x) - 3^x = 0
32^x - 3^x = 0
3*(2^x - 3^(x-1)) = 0
2^x - 3^(x-1) = 0
2^x - 3(2^(x-1)) = 0
2^x - 3(1/2)(2^x) = 0
2^x - 1.5(2^x) = 0
0.5(2^x) = 0
2^x = 0

Since 2^x can never be zero for any real value of x, there is no solution for this case.

2^x - 23^x = 0
2^x = 23^x
(2/3)^x = 1
x = 0

Therefore, the only solution to the equation is x = 0.