SOLUTION: Really confused and need help with this problem: Log(base3)x=Log(base9)6x

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Question 841914: Really confused and need help with this problem:
Log(base3)x=Log(base9)6x
Answer by josgarithmetic(39623) About Me  (Show Source):
You can put this solution on YOUR website!
log%283%2Cx%29=log%289%2C6x%29

Change of Base Formula: highlight_green%28log%28b%2Cx%29=log%28a%2Cx%29%2Flog%28a%2Cb%29%29
The right hand side of the given equation should be converted to base 3.
log%289%2C6x%29=log%283%2C6x%29%2Flog%283%2C9%29
log%289%2C6x%29=%281%2F2%29log%283%2C6x%29
Using that in the original given equation,

log%283%2Cx%29=%281%2F2%29log%283%2C6x%29
2log%283%2Cx%29=log%283%2C6x%29
log%283%2Cx%5E2%29=log%283%2C6x%29

The antilog of left and right can be taken, and the arguments inside the log function are equated, and ....
x%5E2=6x
x%5E2-6x=0
x%28x-6%29=0
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highlight%28x=0%29 or highlight%28x=6%29