Question 324938
I'll assume you meant log base 2:
{{{log(2,x) + log(2,(x - 1)) = 1}}}
{{{log(2,x(x - 1)) = 1}}}
{{{x(x - 1) = 2^1}}}
{{{x^2 - x = 2}}}
{{{x^2 - x - 2 = 0}}}
{{{(x-2)(x+1) = 0}}}
x = {-1, 2}
.
But wait, -1 is an extraneous solution toss it out.  Leaving you with:
x = 2