SOLUTION: Solve the following expression using a common base: (1/2)^(4x+1)=(1/4)^(3x+3) I figured that I could turn 1/2 into (1/4)^2, (you end up with x=1/5), but LS does not equal RS wit

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: Solve the following expression using a common base: (1/2)^(4x+1)=(1/4)^(3x+3) I figured that I could turn 1/2 into (1/4)^2, (you end up with x=1/5), but LS does not equal RS wit      Log On


   

Question 408847: Solve the following expression using a common base:
(1/2)^(4x+1)=(1/4)^(3x+3)
I figured that I could turn 1/2 into (1/4)^2, (you end up with x=1/5), but LS does not equal RS with x=1/5
here's what I tried:
(1/2)^(4x+1)=(1/4)^(3x+3)
(1/4)^2(4x+1)=(1/4)^(3x+3)
(1/4)^(8x+2)=(1/4)^(3x+3)
8x+2=3x+3
8x-3x=3-2
5x=1
x=(1/5)
LS:
=(1/2)^(4x+1)
=(1/2)^(4(1/5)+1)
=(1/2)^1.8
=0.287
RS:
=(1/4)^(3x+3)
=(1/4)^(3(1/5)+3)
=(1/4)^(3.6)
=0.0068
LS does not equal RS, so I'm oviously wrong...
I'd truly appreciate any help I could get!! Thank You :)
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!


Hi

Solving the following expression using a 'common base':

(1/2)^(4x+1)=(1/4)^(3x+3) 

(1/2)^(4x+1)=(1/2)^2(3x+3)

     4x+ 1 = 2(3x+3)

     4x+ 1 = 6x+6

         -5 = 2x

        -5/2 = x 

CHECKING our Answer***

(1/2)^-9 = (1/4)^(-9/2)= (1/2)^-9