SOLUTION: Solve: log(x-3) + log(x+5) =1 My teacher said to use the log of products to get an equivalent quadratic equation. I don't really follow him on how to do it, though.

Click here to see ALL problems on logarithm

Question 299113: Solve: log(x-3) + log(x+5) =1
My teacher said to use the log of products to get an equivalent quadratic equation. I don't really follow him on how to do it, though.
Answer by nerdybill(7384) (Show Source):
You can put this solution on YOUR website!
What he means is to apply a "log rule":
log a + log b = log (ab)
.
Solve: log(x-3) + log(x+5) =1
log(x-3)(x+5) =1
(x-3)(x+5) =10^1
x^2+5x-3x-15 = 10
x^2+2x-15 = 10
x^2+2x-25 = 0
Solve by applying the quadratic formula. Doing so yields:
x = {4.1, -6.1}
Plug the above back into the original equation to check whether they are viable answers. The negative answer will not work (it's an extraneous solution) throw it out leaving:
x = 4.1
.
Details of quadratic formula to follow:

Solved by pluggable solver: SOLVE quadratic equation with variable

Quadratic equation (in our case ) has the following solutons:

For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=104 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: 4.09901951359278, -6.09901951359278. Here's your graph: