SOLUTION: Please help me solve this problem: log2 (x-1)+ log2 (x+3)=3

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Question 734070: Please help me solve this problem:
log2 (x-1)+ log2 (x+3)=3
Found 2 solutions by lwsshak3, nerdybill:
Answer by lwsshak3(11628) (Show Source):
You can put this solution on YOUR website!
Please help me solve this problem:
log2 (x-1)+ log2 (x+3)=3
log2[ (x-1)(x+3)]=3
log2[(x-1)(x+3)]=log2(8)
(x-1)(x+3)=8
x^2+2x-3=8
x^2+2x-11=0
solve for x by quadratic formula:
x≈-4.4641 (reject, (x+3)>0)
or
x≈2.4641

Answer by nerdybill(7384) (Show Source):
You can put this solution on YOUR website!
log2 (x-1)+ log2 (x+3)=3
log2 ((x-1)(x+3)) = 3
(x-1)(x+3) = 2^3
x^2+3x-x-3 = 8
x^2+2x-3 = 8
x^2+2x-11 = 0
applying the "quadratic formula" we get:
x = {2.46, -4.46}
Throw out the negative answer (extraneous) leaving:
x = 2.46
.
Details of quadratic formula follows:

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=48 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: 2.46410161513775, -4.46410161513775. Here's your graph: