SOLUTION: log_2 (x+3)+log_2(x-3)=4 Solve for x

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Question 969493: log_2 (x+3)+log_2(x-3)=4
Solve for x
Found 2 solutions by ankor@dixie-net.com, Boreal:
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
log%282%2C%28x%2B3%29%29+%2B+log%282%2C%28x-3%29%29+=+4
log%282%2C%28x%2B3%29%28x-3%29%29+=+4
FOIL
log%282%2C%28x%5E2-9%29%29+=+4
the exponent equiv
%28x%5E2-9%29+=+2%5E4
x^2 - 9 = 16
x^2 = 16 + 9
x^2 = 25
x = +/-sqrt%2825%29
x = +5, the solution
and
x = -5, will not work in the original problem (log of a neg)


Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
x=5
Start by rewriting as Log (b2) {(x+3)(x-3)}=4. Adding logs is like multiplying them with the same base.
Think of log 10 +log 100=1+2=3; log 10*100=log 1000=3.
From the definitions of logs, 2^4=16=(x+3)(x-3)=x^2-9.
So, x^2-9=16; x^2=25; x+ 5 or -5. Negative logs don't exist, so -5 is extraneous.
Try 5 in the original
Log b2 (8) +log b2 (2)=4 ?
The first is 3 (2^3=8) and the second is 1; 3+1=4.