SOLUTION: hi i have a question about a Logarithm problem log(x-6)-log(x-2)=log x/5 i understand (X-6)/(X-2) = LOG X - LOG 5 ""WHY DOES X GOES WITH (X-6) AND 5 GOES WITH (X-2)""""

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Question 188010: hi i have a question about a Logarithm problem
log(x-6)-log(x-2)=log x/5
i understand (X-6)/(X-2) = LOG X - LOG 5
""WHY DOES X GOES WITH (X-6) AND 5 GOES WITH (X-2)""""
so the problem goes something like this x(x-6)/5(x-2)=10^0 =1
x^2-6x=1*5(x-2) = which is x^2-11x-10=0
Answer by Alan3354(69443)   (Show Source): You can put this solution on YOUR website!
hi i have a question about a Logarithm problem
log(x-6)-log(x-2)=log x/5
i understand (X-6)/(X-2) = LOG X - LOG 5
""WHY DOES X GOES WITH (X-6) AND 5 GOES WITH (X-2)""""
so the problem goes something like this x(x-6)/5(x-2)=10^0 =1
x^2-6x=1*5(x-2) = which is x^2-11x-10=0
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log(x-6)-log(x-2)=log x/5 Solve for x ?
log((x-6)/(x-2)) = log(x/5)
(x-6)/(x-2) = x/5
x*(x-2) = 5*(x-6)


Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation (in our case ) has the following solutons:



For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

The discriminant -71 is less than zero. That means that there are no solutions among real numbers.

If you are a student of advanced school algebra and are aware about imaginary numbers, read on.


In the field of imaginary numbers, the square root of -71 is + or - .

The solution is , or
Here's your graph:

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The logs of complex numbers is outside the scope of this forum.
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