Question 1139555
<br>
{{{log(2,(x))+log(2,(x-3)) = 2}}}<br>
Rewrite the right side in terms of log base 2:<br>
{{{log(2,(x))+log(2,(x-3)) = log(2,4)}}}<br>
Write the left side as a single logarithm using basic rules of logarithms:<br>
{{{log(2,(x(x-3))) = log(2,4)}}}<br>
The logs are equal, so the arguments are equal:<br>
{{{x(x-3) = 4}}}
{{{x^2-3x-4 = 0}}}
{{{(x-4)(x+1) = 0}}}<br>
{{{x = 4}}} or {{{x = -1}}}<br>
x=-1 does not work; the argument of a logarithm can't be negative.<br>
x=4 works:<br>
{{{log(2,4)+log(2,(4-3)) = log(2,4)+log(2,1) = 2+0 = 2}}}<br>
ANSWER: x = 4