How do you solve 4^x=8^(w-1)?

Question

I think I'm supposed to use chunking, but I don't know how.

Answer 1

4^x = 8^(w - 1)

(2^2)^x = (2^3)^(w - 1)

2^(2x) = 2^[3(w -1)]

2x = 3(w - 1)

x = 3(w - 1)/2

Answer 2

i assume you meant:
4^x = 8^(x - 1)

change to the base of 2

4 = 2^2
8 = 2^3

(2^2)^x = (2^3)^(x - 1)

2^(2x) = 2^(3x - 3)

since they are the same base, the exponent must be equal to each other

2x = 3x - 3
-x = -3
x = 3 <== answer

EDIT: if not, then the answerers below me are correct.

Answer 3

first let them have the same base

since 4=2^2
you can write it this way
2^2x

since 8=2^3
yo can write it this way
2^3(w-1)

Now that they have the same base you no longer need to worry about them just set the exponents equal each other.

2x=3(w-1)
then thats just simple algebra

Hope this helps.

Answer 4

x log(4) = (w-1) log(8)
x= (w-1) log(8) / log(4)
This is the case if you are solving for x. You'll have to know w, if you ned a number on the right.
You could also use ln instead of log.

Answer 5

4^x=8^(w-1)
(2^2)^x = (2^3)^(w-1)
2^(2x) =2^(3w-3)
Thus 2x = 3w-3
x = (3w-3)/2