SOLUTION: Why x = 7 is not a solution to the equation: {{{log(10,(x+3)(x-8)) + log(10,((x+3)/(x-8))) = 2}}}

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Question 1083333: Why x = 7 is not a solution to the equation:

Found 2 solutions by josgarithmetic, Theo:
Answer by josgarithmetic(39623) (Show Source):
You can put this solution on YOUR website!
x=7 will make one of the logarithms have a negative input value, and we do not take logs of negative values.

Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website!
log((x+3)/(x-8)) would lead to the log of a negative number which is not allowed.

i believe the equation is log((x+3)*(x-8)) + log((x+3)/(x-8)) = 2

that makes more sense than having the (x-8) outside of the log function as you show it.

i graphed it and found x to be equal to -13.

that value of x makes the equation true.

i then proceeded to solve it algebraically and came up with the source of the problem.

here's what i did:

start with log((x+3)*(x-8)) + log((x+3)/(x-8)) = 2

this is equivalent to log(x+3) + log(x-8) + log(x+3) - log(x-8) = 2

the plus and minus log(x-8) cancel out and you are left with:

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

this is equivalent to 2 * log(x+3) = 2

i then divided both sides of the equation by 2 to get log(x+3) = 1

this is true if and only if 10^1 = x+3 which results in x = 7.

that doesn't work because log((x+3)/(x-8)) becomes the log of a negative number which is not allowed.

i then looked at it in a different way.

starting from 2 * log(x+3)) = 2, i used the fact that 2 * log(x+3)) is equivalent to log((x+3)^2), so the equation became:

log((x+3)^2) = 2

this is true if and only if 10^2 = (x+3)^2

this becomes 100 = (x+3)^2

take the square root of both sides of the equation and you get:

(x+3) = plus or minus 10.

x + 3 = 10 leads to x = 7

x + 3 = -10 leads to x = -13

bingo !!!!!

x = -13 is the answer.

this was confirmed graphically as shown below: