solve: 4^2x - 20(4^x) + 64 = 0?

Question

i know the answer is 2 or 1, but don't know how to get it?, thanks

Answer 1

let's say u=4^x
so the equation becomes:
u^2-20u+64=0
(u-16)(u-4)=0
u=16 or u=4

substitute 4^x in to get:
4^x=16 or 4^x=4

thus, x=2 or x=1

Answer 2

Factor by FOIL

(4^x - 16)(4^x - 4) = 0

So 4^x = 16 (which would give you the 2)

or 4^x = 4 (which gives the 1)

Answer 3

supstitute 4^x with t, then you get t^2 - 20t +64, then find t1 and 2 with the formula t1,2=(20+or- Sqr(400-4*64))/2; you get t1=16, thts 4^2 and t2=4; from that 4^x=4^2 and 4^x=4 you get x1=2, and x2=1

Answer 4

4^2x - 20(4^x) + 64 = 0
(4^x - 16)(4^x - 4)
4^x = 16 or 4
x = log(16)/log(4) = log(4)/log(4)
x = 2 or 1