SOLUTION: Please help to Solve for x: Log (x+3)+Log (x-2)=Log14 (all 3 Logs are to the base 10)

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Question 1058789: Please help to Solve for x:
Log (x+3)+Log (x-2)=Log14
(all 3 Logs are to the base 10)

Found 2 solutions by solve_for_x, Alan3354:
Answer by solve_for_x(190) (Show Source):
You can put this solution on YOUR website!
Using the logarithm law log(a) + log(b) = log(ab) gives:

log((x+3)(x-2)) = log(14)

Taking the antilog of both sides gives:

(x + 3)(x - 2) = 14

x^2 + x - 6 = 14

x^2 + x - 20 = 0

(x +5)(x - 4) = 0

x = -5, x = 4

Since the log function is not defined for values less than 0, the
x = -5 solution is discarded because -5 + 3 = -2 and -5 - 2 = -7.

The only solution that works is x = 4.

Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website!
Log (x+3)+Log (x-2)=Log14
Log((x+3)*(x-2)) = Log14
Log((x+3)*(x-2)) = Log(14)
x^2 + x - 6 = 14
x^2 + x - 20 = 0
(x+5)*(x-4) = 0
x = -5 ** rejected, gives log of a negative value.
x = 4