SOLUTION: how to solve: log(2x+3)+log(x-2)=2logx

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Question 375867: how to solve:
log(2x+3)+log(x-2)=2logx
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
log%28%282x%2B3%29%29%2Blog%28%28x-2%29%29=2log%28%28x%29%29
log%28%282x%2B3%29%28x-2%29%29=log%28%28x%5E2%29%29
%282x%2B3%29%28x-2%29=x%5E2
2x%5E2-x-6=x%5E2
x%5E2-x-6=0
%28x-3%29%28x%2B2%29=0
Two solutions:
x-3=0
x=3
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x%2B2=0
x=-2
Verify the solutions.
log%28%282x%2B3%29%29%2Blog%28%28x-2%29%29=2log%28%28x%29%29
log%28%282%283%29%2B3%29%29%2Blog%28%283-2%29%29=2log%28%283%29%29
log%28%289%29%29%2Blog%28%281%29%29=2log%28%283%29%29
log%28%289%29%29=log%28%289%29%29
.
.
.
log%28%282x%2B3%29%29%2Blog%28%28x-2%29%29=2log%28%28x%29%29
log%28%282%28-2%29%2B3%29%29%2Blog%28%28-2-2%29%29=2log%28%28-2%29%29
This solution is not allowed since the log function requires a positive argument.
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highlight%28x=3%29