SOLUTION: Solving for x when... 16^x = (1/2)^2*(8^(2x-1)) So far I've done... 16^x = (1/4)(8^(2x-1)) log(16^x) = log(2^(2x-1)) xlog(16) = (2x-1)*log(2) That's where I stopped.. This i

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: Solving for x when... 16^x = (1/2)^2*(8^(2x-1)) So far I've done... 16^x = (1/4)(8^(2x-1)) log(16^x) = log(2^(2x-1)) xlog(16) = (2x-1)*log(2) That's where I stopped.. This i      Log On


   

Question 1065315: Solving for x when...
16^x = (1/2)^2*(8^(2x-1))
So far I've done...
16^x = (1/4)(8^(2x-1))
log(16^x) = log(2^(2x-1))
xlog(16) = (2x-1)*log(2)
That's where I stopped..
This is for a college population dynamics class -- we're doing algebra puzzles (for some reason).
Answer by josgarithmetic(39620) About Me  (Show Source):
You can put this solution on YOUR website!
HINT: Put everything into base-two.

%282%5E4%29%5Ex=%282%5E%28-1%29%29%5E2%2A%28%282%5E3%29%5E%282x-1%29%29

2%5E%284x%29=%282%5E%28-2%29%29%2A%282%5E%283%282x-1%29%29%29

2%5E%284x%29=2%5E%28-2%2B3%282x-1%29%29
The same base on both sides raised to the powers shown means the powers must be equal.

4x=-2%2B3%282x-1%29
Continue to solve for x.