Question 1083333
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:


<img src = "http://theo.x10hosting.com/2017/053101.jpg" alt="###" </>