SOLUTION: log2 (x-3)+log2 x-log2 (x+2)=2

Question 420445: log2 (x-3)+log2 x-log2 (x+2)=2
Answer by nerdybill(7384) (Show Source): You can put this solution on YOUR website!
log2 (x-3)+log2 x-log2 (x+2)=2
log2 x(x-3) -log2 (x+2)=2
log2 x(x-3)/(x+2)=2
x(x-3)/(x+2)=2^2
x(x-3)/(x+2)=4
x(x-3)=4(x+2)
x^2-3x=4x+8
x^2-7x=8
x^2-7x-8=0
(x-8)(x+1) = 0
x = {8, -1}
throw out the -1 -- extraneous
leaving
x = 8

RELATED QUESTIONS
Log2 (x+2) = log2 3-log2... (answered by Alan3354,MathTherapy)

Log2... (answered by benazir.sj@gmail.com)

Log2^(x-3)=2-log2^x (answered by jsmallt9)

log2(x+2)-log2(x)=3 (answered by ikleyn)

log2(x)+log2(x-3)=2 (answered by josgarithmetic,greenestamps)

3 log2 3+2 log2... (answered by jim_thompson5910)

Solve: Log2(2) +log2(x+3)=5 (answered by stanbon)

log2(x+2)=2+log2(x-2) (answered by josmiceli)

log2(x+2)+log2(x-2)=5 (answered by robertb)