Question 1198977

solve for {{{x}}}: 

{{{1+log(2,x^2-4x-16)=log(2,x^2-3x+4)}}}

{{{1=log(2,x^2-3x+4)-log(2,x^2-4x-16)}}}

{{{1=log(2,(x^2-3x+4)/(x^2-4x-16))}}}..........{{{log(2,2)=1}}}

{{{log(2,2)=log(2,(x^2-3x+4)/(x^2-4x-16))}}}

{{{2=(x^2-3x+4)/(x^2-4x-16)}}}

{{{2(x^2-4x-16)=(x^2-3x+4)}}}

{{{2x^2-8x-32=x^2-3x+4}}}

{{{2x^2-8x-32-x^2+3x-4=0}}}

{{{x^2-5x-36=0}}}..........use quadratic formula

{{{x=(-(-5)+-sqrt((-5)^2-4*1*(-36)))/(2*1)}}}

{{{x=(5+-sqrt(25+144))/2}}}

{{{x=(5+-sqrt(169))/2}}}

{{{x=(5+-13)/2}}}

solutions:

{{{x=(5+13)/2}}}=>{{{x=9}}}

{{{x=(5-13)/2}}}=>{{{x=-4}}}