SOLUTION: log2(x - 6) + log2(x - 4) = log2 x solve the logarithmic equation. and round.

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Question 883161: log2(x - 6) + log2(x - 4) = log2 x solve the logarithmic equation. and round.
Found 2 solutions by lwsshak3, MathTherapy:
Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
log2(x - 6) + log2(x - 4) = log2 x solve the logarithmic equation. and round.
***
solve for x
log%282%2C%28x-6%29%29%2Blog%282%2C%28x-4%29%29=log%282%2C%28x%29%29
log%282%2C%28x-6%29%28x-4%29%29=log%282%2C%28x%29%29%29
x=(x-6)(x-4)
x=(x^2-10x+24)
x^2-11x+24=0
(x-3)(x-8)=0
x=3
or
x=8

Answer by MathTherapy(10801) About Me  (Show Source):
You can put this solution on YOUR website!
log2(x - 6) + log2(x - 4) = log2 x solve the logarithmic equation. and round. 
****************************************************************************
x = 3 or x = 8, as stated by the other person, is partly WRONG!   

log+%282%2C+%28x+-+6%29%29+%2B+log+%282%2C+%28x+-+4%29%29+=+log+%282%2C+%28x%29%29
Looking at the logarithmic equation, the smallest log argument, x - 6 MUST be > 0.
So, x - 6 > 0, and therefore, x > 6. 

We then have: log+%282%2C+%28x+-+6%29%29+%2B+log+%282%2C+%28x+-+4%29%29+=+log+%282%2C+%28x%29%29, with x > 6

The x-value 3 is NOT > 6, which makes it EXTRANEOUS and an INVALID/UNACCEPTABLE solution. On the
other hand, the x-value 8 is > 6, which makes it the ONLY solution to this equation.