SOLUTION: Hi, I am stuck log (x+6)-log(x+2)=log(x) Does log(x) become 1, because 10^0? Thank you in advance

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Question 246565: Hi,
I am stuck
log (x+6)-log(x+2)=log(x)
Does log(x) become 1, because 10^0?
Thank you in advance



Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!

log+%28%28x%2B6%29%29-log%28%28x%2B2%29%29=log%28%28x%29%29

Use the rule log%28%28A%29%29-log%28%28B%29%29=log%28%28A%2FB%29%29 on the left side:

log%28%28%28x%2B6%29%2F%28x%2B2%29%29%29=log%28%28x%29%29

What the log is taken of on the left side must equal to
what the log is taken of on the right side.

%28%28x%2B6%29%29%2F%28%28x%2B2%29%29=x

Multiply both sides by %28x%2B2%29

%28%28x%2B6%29%2F%28x%2B2%29%29%28x%2B2%29=x%28x%2B2%29

%28%28x%2B6%29%2F%28cross%28x%2B2%29%29%29%28cross%28x%2B2%29%29=x%28x%2B2%29

x%2B6=x%5E2%2B2x

0=x%5E2%2Bx-6

x%5E2%2Bx-6=0

%28x%2B3%29%28x-2%29=0

x%2B3=0       x-2=0
x=-3         x=2

The negative solution must be discarded
because it cannot be substituted in the 
term on the right of the original equation,
log%28%28x%29%29 for the log of a negative
number is not a real number.

Therefore the only solution is x=2

Edwin