SOLUTION: Solve log2(x-2)-log2(x+5)=3

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Question 954792: Solve log2(x-2)-log2(x+5)=3
Found 2 solutions by MathLover1, MathTherapy:
Answer by MathLover1(20855) About Me  (Show Source):
You can put this solution on YOUR website!

log%282%2C%28x-2%29%29-log%282%2C%28x%2B5%29%29=3
log%282%2C%28x-2%29%2F%28x%2B5%29%29=3....change to base 10
log%28%28x-2%29%2F%28x%2B5%29%29%2Flog%282%29=3
log%28%28x-2%29%2F%28x%2B5%29%29=3log%282%29
log%28%28x-2%29%2F%28x%2B5%29%29=log%282%5E3%29
log%28%28x-2%29%2F%28x%2B5%29%29=log%288%29.....if log same, then
%28x-2%29%2F%28x%2B5%29=8
%28x-2%29=8%28x%2B5%29
x-2=8x%2B40
-40-2=8x-x
-42=7x
-42%2F7=x
x=-6

Answer by MathTherapy(10801) About Me  (Show Source):
You can put this solution on YOUR website!
Solve log2(x-2)-log2(x+5)=3
*****
The solution, x = - 6, by the other person who responded, is WRONG!! 

log+%282%2C+%28x+-+2%29%29+-+log+%282%2C+%28x+%2B+5%29%29+=+3
The SMALLER of the 2 log arguments, x - 2 MUST be > 0. So, x - 2 > 0 ===> x > 2.
We then have: log+%282%2C+%28x+-+2%29%29+-+log+%282%2C+%28x+%2B+5%29%29+=+3, with x > 2.
                                           log+%282%2C+%28%28x+-+2%29%2F%28x+%2B+5%29%29%29+=+3 ----- Applying log+%28b%2C+%28c%29%29+-+log+%28b%2C+d%29%29 = log+%28b%2C+%28c%2Fd%29%29
                                                       %28x+-+2%29%2F%28x+%2B+5%29+=+2%5E3 ---- Converting to EXPONENTIAL form 
                                                       %28x+-+2%29%2F%28x+%2B+5%29+=+8
                                                   8(x + 5) = x - 2 ----- Cross-multiplying 
                                                    8x + 40 = x - 2 
                                                        8x - x = - 2 - 40
                                                              7x = - 42                                                                                                 
                                                                x+=+%28-+42%29%2F7+=+-+6 
                                              
The x-value, - 6, is NOT > 2, and is therefore an EXTRANEOUS solution, which makes it an INVALID/UNACCEPTABLE solution. 
So, this equation DOESN'T have a solution!!