Question 537901
Solve for x:
{{{2Log[2](x)-Log[2](x-3) = 2+Log[2](4)}}} Apply the "power rule" to the first term on the left side.
{{{Log[2](x^2)-Log[2](x-3) = 2+Log[2](4)}}} Apply the "quotient rule" to the left side.
{{{Log[2](x^2/(x-3)) = 2+Log[2](4)}}} Substitute {{{Log[2](4) = 2}}}
{{{Log[2](x^2/(x-3)) = 2+2}}} 
{{{Log[2](x^2/(x-3)) = 4}}} Rewrite this in exponential form:
{{{2^4 = x^2/(x-3)}}} Substitute {{{2^4 = 16}}}
{{{16 = x^2/(x-3)}}} Multiply both sides by {{{(x-3)}}}
{{{16(x-3) = x^2}}} Simplify.
{{{16x-48 = x^2}}} Rewrite as:
{{{x^2-16x+48 = 0}}} Solve by factoring.
{{{(x-4)(x-12) = 0}}} Apply the "zero product rule".
{{{x-4 = 0}}} or {{{x-12 = 0}}} therefore:
{{{highlight(x = 4)}}} or {{{highlight_green(x = 12)}}}