Given that:
27^x + 64^x 193
----------- = ----
48^x + 36^x 144
Find all possible real values of x, that satisfy the given conditions.
I will start by reducing the left side a bit. Rewrite the bases in terms of 3 and 4:
[(3^x)^3 + (4^x)^3] / [(3^x)*(4^x)^2 + (3^x)^2*(4^x)]
Now it is more readily apparent on how to factor. Factoring and canceling the common factor leaves:
[(3^x)^2 - (3^x)*(4^x) + (4^x)^2] / [(3^x)*(4^x)]
Then distribute the fraction through and simplify:
(3/4)^x + (4/3)^x - 1
Substitute this back, then given equation reduces to:
(3/4)^x + (4/3)^x = 337/144
Now I will multiply through by (4/3)^x to create a quadratic:
[(4/3)^x]^2 - (337/144)*[(4/3)^x] + 1 = 0
The quadratic formula, plus some arithmetic eventually yields:
(4/3)^x = 16/9 or 9/16.
Then it is easy to see x=2 or -2 are the two solutions.
Solution
