Question 169365
Please help me solve {{{log 12 ^ (2x) + log 12 ^ (x-1) = 1}}}.

I have this so far: {{{log 12 (2x * x-1) = 1}}}.
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It's {{{log(12,(2x^2-2x)) = 1}}}  Multiply the 2x times the 1, also
1 is {{{log(12,12)}}}, so
{{{log 12 (2x^2-2x) = log(12,12)}}}
Then, 
{{{2x^2 - 2x = 12}}}
{{{x^2 - x - 6 = 0}}}
(x-3)*(x+2) = 0
x = 3
x = -2