Question 173595: Log(base6)x+log(base6)(x-5)=2
Check for inadmissible roots
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! Log(base6)x+log(base6)(x-5)=2
:
log6(x) + log6(x-5) = 2
:
adding logs mean multiply, so we can write it:
log6(x(x-5)) = 2
:
Write it in the exponential form
x(x-5) = 6^2
:
x^2 - 5x = 36
A quadratic equation
x^2 - 5x - 36 = 0
Factor
(x-9)(x+4) = 0
two solutions
x = -4, not an admissible solution (can't have a log of a neg number)
and
x = +9, a good solution
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