How do you solve 4^(x+4) = 8^x4x+4=8x?

1 Answer
Jul 23, 2015

I found: x=8x=8

Explanation:

You can write it as:
(2^2)^(x+4)=(2^3)^x(22)x+4=(23)x
2^(2(x+4))=2^(3x)22(x+4)=23x take the log_2log2 on both sides to get rid of the 22 (where you have cancel(log_2)(cancel(2)^x)=x) and get:
2(x+4)=3x
2x+8=3x
x=8