SOLUTION: please help! i"m trying to solve: log 2 (x-1) - log 2 (x+3) = log 2 (1/x)

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Question 377694:
please help! i"m trying to solve:

log 2 (x-1) - log 2 (x+3) = log 2 (1/x)

This is what i tried, but i think it's wrong: used quotient rule like this: log 2 (x-1/x+3) = log 2 (1/x) OK! NOW WHAT???!!! I'm lost! :(
Found 2 solutions by Fombitz, ewatrrr:
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
You're almost there.
log%282%2C%28%28x-1%29%2F%28x%2B3%29%29%29=log%282%2C%281%2Fx%29%29
%28x-1%29%2F%28x%2B3%29=1%2Fx
Cross multiply,
x%28x-1%29=x%2B3
x%5E2-x=x%2B3
x%5E2-2x-3=0
%28x-3%29%28x%2B1%29=0
Two solutions but only positive arguments are allowed for the log function.
x-3=0
highlight%28x=3%29

Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!


Hi

log 2 (x-1) - log 2 (x+3) = log 2 (1/x)

log 2 (x-1) - log 2 (x+3) - log 2 (1/x)=0



log_2%28%28x-1%29%2F%28x%2B3%29%281%2Fx%29%29=0



2^0 = (x-1)/(x+3)(1/x)

1+=+%28x-1%29%2F%28x%2B3%29%281%2Fx%29

1 = x(x-1)/(x+3)

(x+3) = x^2 -x

x^2 -2x - 3 = 0

(x-3)(x+1)= 0

(x-3)=0    x = 3

(x+1)= 0   x = -1 reject as a solution

Any solution that makes an argument (or base) zero or negative must be rejected.

log 2 ((-1)-1) - log 2 ((-1)+3) = log 2 (1/(-1))