Question 34699: I am having a problem solving this problem, #33 in Dr Rapalje's book, "Stepping through college algebra, one step at a time." log b (x-6)- log b (x)= log b (x-4). I have tried log b x-6 over x-4 = x, x-6 = x(x-4), x-6= x^2-4x, set to 0=x^2-4x-x+6, 0=x^2-5x+6, 0=(x-3) (x-2), x=3,x=2 is the answer I got, but it doesn't match the answer in the back of the book, which is 0, reject 3,2. Thank you.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! log(x-6)/x]= log(x-4)
Take the anti-log of both sides to get:
[(x-6)/x)]=x-4
Cross multiply to get:
x-6=x(x-4)
x-6=x^2-4x
x^2-5x+6=0
(x-3)(x-2)=0
x=3 or x=2
Check them out: neither works because you get negative values in the original where they cannot be in the original equation.
The 0 in your answer book might mean 0 solutions.
0 certainly is not a solution.
Cheers,
Stan H.
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