Question 681634
How would you solve log base 6 of 2x + log base 6 of (x-2) = 1?
Our teacher said the answer should be 3.

So far I condensed it to log base 6 of (2x(x-2)) or log base 6 of (2x^2-4x) = 1
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log base 6 of (2x^2-4x) = 1
--> (2x^2-4x) = 6
x^2 - 2x -3 = 0
(x-3)*(x+1) = 0
x = -1 is rejected since {{{log(6,-3)}}} won't work.
x = 3