Interesting Math Problem - Solve 4^x + 6^x = 9^x?

Question

A few days back some else posted this problem. I would like to see an eloquent solution of same.

Answer 1

Divide both sides by 9^x

(4/9)^x + (6/9)^x = 1

Regroup
[(2/3)^x]^2 + [(2/3)^x] - 1 = 0

Set A = (2/3)^x
so
A^2 + A - 1 = 0
Quadratic Formula
A = [-1 + sqrt(1 + 4)] / 2

So (2/3)^x = [-1 + sqrt(5)] / 2

Take Natural Logarithm of both sides
ln {(2/3)^x} = ln { [-1 + sqrt(5)] / 2 }
x * ln(2/3) = ln { [-1 + sqrt(5)] / 2 }
x = ln{ [sqrt(5) - 1 ] / 2 } / ln (2/3)

Note: the previous posting used "Phi" = (1 + sqrt(5))/2
But this is how I would do it without that knowledge,
which I was not familiar with.

Answer 2

not sure if this is right but here's how I did it.

4^x + 6^x = 9^x

10^2x = 9^x
-8^x =-8^x

2^x = 1^x

2^x = 1

therefore x = 0 any number raise to the power of 0 = 1

Answer 3

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Answer 4

Have you solved it yet? I'm having difficulties and wonder if it CAN be solved...or if I'm just being stupid and forgetting some algebra rules.