Question 677886

{{{log(2,(x^2+5x-32))=2}}}......in base {{{10}}} it will be


{{{log((x^2+5x-32))/log((2))=2}}}


{{{log((x^2+5x-32))=2log((2))}}}.......since {{{2log(2)=log(2^2)=log(4)}}}, we have


{{{log((x^2+5x-32))=log((4))}}}


{{{log((x^2+5x-32))=log((4))}}}..if this equal, than


{{{x^2+5x-32=4}}}...is equal too

now solve for {{{x}}} using quadratic formula


{{{x^2+5x-32-4=0}}}


{{{x^2+5x-36=0}}}


 {{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}


 {{{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 = 8/2 }}}


 {{{x =4 }}}


and


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


 {{{x =-18/2 }}}


 {{{x =-9 }}}