SOLUTION: log3(x-2)+log3(x+4)=3

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Question 166689: log3(x-2)+log3(x+4)=3
Found 2 solutions by nerdybill, MathTherapy:
Answer by nerdybill(7384) About Me  (Show Source):
You can put this solution on YOUR website!
log3(x-2)+log3(x+4)=3
log3[(x-2)(x+4)]=3
(x-2)(x+4)=3^3
x^2+4x-2x-8 = 9
x^2+2x-8 = 9
x^2+2x-17 = 0
.
Since we can't factor, we must use the quadratic equation.
Do so will yield:
x = {3.243, -5.243}
.
Details here:
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B2x%2B-17+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%282%29%5E2-4%2A1%2A-17=72.

Discriminant d=72 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-2%2B-sqrt%28+72+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%282%29%2Bsqrt%28+72+%29%29%2F2%5C1+=+3.24264068711928
x%5B2%5D+=+%28-%282%29-sqrt%28+72+%29%29%2F2%5C1+=+-5.24264068711928

Quadratic expression 1x%5E2%2B2x%2B-17 can be factored:
1x%5E2%2B2x%2B-17+=+1%28x-3.24264068711928%29%2A%28x--5.24264068711928%29
Again, the answer is: 3.24264068711928, -5.24264068711928. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B2%2Ax%2B-17+%29

Answer by MathTherapy(10801) About Me  (Show Source):
You can put this solution on YOUR website!
log3(x-2)+log3(x+4)=3
*********************

The other person who responded is WRONG!! His solutions, "x = {3.243, -5.243}" are WRONG.
If he'd checked his answers, he would've realized this.

The SMALLER log argument, x - 2 MUST be > 0, so x - 2 > 0, and we get: x > 2. We now have:
log+%283%2C+%28x+-+2%29%29+%2B+log+%283%2C+%28x+%2B+4%29%29+=+3, with x being > 2.
        log+%283%2C+%28x+-+2%29%28x+%2B+4%29%29+=+3 ----- Applying log+%28b%2C+%28c%29%29+%2B+log+%28b%2C+%28d%29%29 = log+%28b%2C+%28c%2Ad%29%29
              %28x+-+2%29%28x+%2B+4%29+=+3%5E3 ---- Converting to EXPONENTIAL form
                 x%5E2+%2B+2x+-+8+=+27
           x%5E2+%2B+2x+-+8+-+27+=+0%29
               x%5E2+%2B+2x+-+35+=+0
       (x - 5)(x + 7) = 0
        x - 5 = 0        OR        x + 7 = 0
              x = 5        OR               x = - 7 (ignore)

The constraint above "states" that x MUST be > 2. 5 is > 2, but - 7 is NOT. This makes - 7 an
EXTRANEOUS solution, and the only ACCEPTABLE solution, x  = 5.

You can do the CHECK!!