Question

how do you solve 4^(x+5)= (1/8)^(2x-8)

Answer 1

Take log to the base 2 on both sides.

Answer 2

what would the result be? My teacher says to put 1/8 to 4^-1.5 and solve but I don't know how

Answer 3

Is this your equation?\[4^{(x+5)}=\frac{ 1 }{ 8 }^{(2x-8)}\]

Answer 4

You can solve it without logs also.
The base on the left hand side is 4.
The base on the right hand side is 1/8.
Make both sides have base 2.
4 = 2^2
1/8 = 1/(2^3) = 2^(-3)
\[ \large
4^{x+5}= \left(\frac 18 \right)^{2x-8} \\ \large
(2^2)^{x+5} = (2^{-3})^{2x-8} \\ \large
2^{2x+10} = 2^{-6x+24} \\
2x + 10 = -6x + 24 \\
8x = 14 \\
x = 14/8 = 7/4
\]

Answer 5

Yup!\[x = \frac{ 7 }{ 4 }\]

Answer 6

Ok I didn't realize that you went down to 2 on the fraction and i ddin't know you could mulitply the power of (2^2) but thanks!

Answer 7

You are welcome.