Solve? #log(x-3)+log(x-4)-log(x+5)=0#

1 Answer
Sean · MathFact-orials.blogspot.com
Dec 5, 2017

#x=7#

Explanation:

#log(x-3)+log(x-4)-log(x+5)=0#

We have the following rules of logs:

#Log(uxxv)=logu+logv#

#log(u/v)=logu-logv#

Therefore, we can rewrite our equation:

#log(x-3)(x-4)-log(x+5)=log(((x-3)(x-4))/(x+5))=0#

We know that:

#log1=0#

Therefore,

#((x-3)(x-4))/(x+5)=1#

Multiply both sides by #(x+5)#:

#(x-3)(x-4)=x+5#

#x^2-7x+12-x-5=0#

#x^2-8x+7=0#

#(x-7)(x-1)=0#

#x-7=0#, then #x=7#

#x-1=0#, then #x=1#

#x=1# is not a valid answer because plugging it into the problem equation gives us the first and second term as logs of negative numbers which we can not have.

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