SOLUTION: Log(base6)x+log(base6)(x-5)=2 Check for inadmissible roots

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Question 173594: Log(base6)x+log(base6)(x-5)=2
Check for inadmissible roots
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
log%286%2Cx%29+%2B+log%286%2C%28x-5%29%29+=+2
log%286%2Cx%2A%28x-5%29%29+=+2
6%5E2+=+x%5E2+-+5x
x%5E2+-+5x+-+36+=+0
Use complete the square method
x%5E2+-+5x+%2B+%285%2F2%29%5E2+=+36+%2B+%285%2F2%29%5E2
%28x+-+5%2F2%29%5E2+=+%28144+%2B+25%29+%2F+4
%28x+-+5%2F2%29%5E2+=++169%2F4
x+-+5%2F2+=+13%2F2
x+=+9
and
x+-+5%2F2+=+-%2813%2F2%29
x+=+-4
This negative root is inadmissable because putting it
back into original equation says that a log to base 6
can give a negative result, which is impossible.