How do you solve this equation: 2^(2x)-12(2^x)+32=0?

Answer 1

if you let u = 2^x
then the eq is
u^2 -12u + 32 = 0
(u-8)(u-4) = 0

u = 8 2^x = 8 x = 3
u = 4 2^x = 4 x = 2

Answer 2

a^2-12a+32=0 a^2 - 8a - 4a + 32 = 0 [ because of the fact all of us understand that -12a = -8a -4a, additionally all of us understand 8*4 = 32, it extremely is final term ...you may constantly look for this accepted in factoring ] a(a - 8) -4(a - 8) = 0 [ Taking difficulty-loose factors out ] (a - 8)(a -4) = 0 [ Multiplicative distribution over addition ] So the two (a -8) = 0 wherein case a = 8, or (a -4 ) = 0, the place a = 4 .... So a = 8 or 4

Answer 3

2^2x-12^2x+32=0, so

(2^x-8)(2^x-4) = 0, so

2^x = 8, or 2^x = 4, so

2^3 = 8, or 2^2 = 4, so

x = 3, or x = 2.

Answer 4

4(2^x)-12(2^x)+32=0
-8(2^x)+32=0
32=8(2^x)
32/8=2^x
4=2^x
2=X
x=2

Answer 5

quadratic formula, or put it in a graphing calculator.