Quadratic equations: variables in exponents using prime numbers

Question

I can't find the solution, could I get some help please?

2^2x - 24.2^x + 128 = 0

^{thanks for the help}

Answer 1

We can solve like this:

First call 2^x = y and note that 2^2x = (2^x)², we can write the equation as follows:

y² - 24y + 128 = 0

y² - 24y + 144 - 16 = 0

(y-12)² = 16

y -12 = 4 or y-12 = -4

y = 16 or y=8

Remembering y = 2^x:

If y = 16, then x = 4

If y = 8, then x = 3

x = 3 or 4