SOLUTION: We have tried and tried to find the answer, but can't seem to figure out when to use the multipicative inverse to solve it. The problem is:
2^2x = 8^(x-1) Hope you can help. Tha
Algebra ->
Exponents-negative-and-fractional
-> SOLUTION: We have tried and tried to find the answer, but can't seem to figure out when to use the multipicative inverse to solve it. The problem is:
2^2x = 8^(x-1) Hope you can help. Tha
Log On
Question 101910: We have tried and tried to find the answer, but can't seem to figure out when to use the multipicative inverse to solve it. The problem is:
2^2x = 8^(x-1) Hope you can help. Thanks! Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! The problem is:
2^2x = 8^(x-1)
------------------
Write both sides to the base 2 as follows:
2^(2x) = (2^3)^(x-1)
2^2x = 2^[3x-3]
Since the bases are the same the exponents are equal.
2x = 3x-3
-x = -3
x = 3
===========
Cheers,
Stan H.
(