solve log2 (6x-8) - log2 (2x+4) - log2 (x-6)?

a solution is log2(3x-4) - log2(x+2) - log2(x-6)

equivalent to log2( ( (3x-4) / (x+2) ) / (x-6 ) )

If we assume that the final minus sign is an = sign and then apply the logarithm rule for quotients, the equation simplifies to the following: log2 ((6x-8)/(2x+4)) = log2 (x-6). But the logarithmic functions are 1-1, so (6x-8)/((2x+4) = x-6. Now multiply both sides of this equation by 2x+4 yielding the result 2x^2 - 8x - 24 = 6x - 8, and write this quadratic equation in standard form (descending powers of independent variable x): 2x^2 - 14x - 16 = 0. Divide both sides of the equation by 2: x^2 - 7x -8 = 0. Factor the simplified equation: (x-8)(x+1) = 0. Set each factor equal to 0 and solve: x = 8 or x = -1. Since x = -1 would require that we take the log of a negative number in the original equation, we reject it as an extraneous root, and the solution set contains only the result x = 8, which we readily see satisfies the original equation.

please check your question... did u misstype an equal sign ...

i cant solve for x , if im not given any details as to what it equals to...

Hi there - an equation must have an equality sign! Where is it?