SOLUTION: How would you solve: 4^(x+1)=16^x

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: How would you solve: 4^(x+1)=16^x      Log On


   

Question 1080320: How would you solve:
4^(x+1)=16^x
Found 2 solutions by josgarithmetic, ikleyn:
Answer by josgarithmetic(39630) About Me  (Show Source):
Answer by ikleyn(52884) About Me  (Show Source):
You can put this solution on YOUR website!
.
4%5E%28x%2B1%29 = 16%5Ex   is the same as   4%5E%28x%2B1%29 = 4%5E%282x%29.


Hence x+1 = 2x,  which implies   x = 1.


Check it on your own.

Solved.