Question 457945
{{{log(2, x) + log(2, 2x+1) = 0}}}

==> {{{log(2, x(2x+1)) = 0}}}

==> {{{x(2x+1) = 2^0}}}

==> {{{2x^2 + x - 1 = 0}}}

==> (2x-1)(x+1) = 0 ==> x = 1/2, -1.

Reject x = -1 (will not satisfy the original equation).

Hence the final answer is x = 1/2.