Question 1198977
<br>
{{{1+log(2,(x^2-4x-16))=log(2,(x^2-3x+4))}}}<br>
Rewrite the first term on the left as a base 2 logarithm.<br>
{{{log(2,(2))+log(2,(x^2-4x-16))=log(2,(x^2-3x+4))}}}<br>
Use the rule for adding logarithms with the same base.<br>
{{{log(2,(2(x^2-4x-16)))=log(2,(x^2-3x+4))}}}<br>
{{{log(2,(2x^2-8x-32))=log(2,(x^2-3x+4))}}}<br>
Set the arguments equal to each other and solve the resulting quadratic equation.<br>
{{{2x^2-8x-32=x^2-3x+4}}}
{{{x^2-5x-36=0}}}
{{{(x-9)(x+4)=0}}}
{{{x=9}}} or {{{x=-4}}}<br>
Both values satisfy the original equation, so<br>
ANSWERS: x = 9 or x = -4<br>