SOLUTION: I need help solving for x in the following equation log(x+3) + log(x-1) = log 5

Question 473707: I need help solving for x in the following equation log(x+3) + log(x-1) = log 5
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
equation is:
log(x+3) + log(x-1) = log(5)
since log(x) + log(y) = log(x*y), this equation becomes:
log((x+3)*(x-1)) = log(5)
this will be true if (x+3)*(x-1) = 5
multiply the factors together to get:
x^2 + 3x - x - 3 = 5
this becomes x^2 + 2x - 8 = 0
this can be factored to get:
(x+4)*(x-2) = 0
this leads to:
x = -4 or x = 2.
when x = -4, the original equation becomes:
log(-1) + log(-5) = log(5)
those logs don't work because you can't get the log of a negative number.
however, if you use the laws of logarithms to combine the 2 logs on the left to one log, then you get:
log((-1)*(-5)) = log(5) which becomes log(5) = log(5) which is true.
when x = 2, the original equation becomes:
log(5) + log(1) = log(5)
combine the logs on the left hand side and you get:
log(5*1) = log(5) which becomes log(5) = log(5) which is true.
the solution to this equation is:
x = -4 or x = 2.
there is that funny little wrinkle where:
log(-1) + log(-5) = log(5) is not valid, but log((-1)*(-5)) = log(5) is valid.

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