SOLUTION: HI I have a question about this equation 4^(6+x)=(1/2)^(7-x).I know the answer to this equation is -19, but I do not know how to get that. Please help me with the steps

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: HI I have a question about this equation 4^(6+x)=(1/2)^(7-x).I know the answer to this equation is -19, but I do not know how to get that. Please help me with the steps      Log On


   

Question 867515: HI I have a question about this equation 4^(6+x)=(1/2)^(7-x).I know the answer to this equation is -19, but I do not know how to get that. Please help me with the steps
Found 2 solutions by josgarithmetic, lwsshak3:
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
Change both expressions to use a base of 2.

The equation starts its first step as,
%282%5E2%29%5E%286%2Bx%29=%282%5E-1%29%5E%287-x%29
Second step can be,
2%5E%282%286%2Bx%29%29=2%5E%28-%287-x%29%29

Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
4^(6+x)=(1/2)^(7-x)
Whenever you see variables in exponents, it usually is a good idea to solve problem with logarithms.
(6+x) log4=(7-x)log(1/2)
(6+x)/(7-x)=log(1/2)/log4
(6+x)/(7-x)=-0.5
6+x=-.5(7-x)
6+x=-3.5+.5x
.5x=-9.5
x=-19