Question 1138795

{{{4log(16,x) = log(4,x + 12)}}}.............change to base {{{10}}}


{{{4log(x)/log(16) = log(x + 12)/log(4)}}}


{{{log(x^4)/log(4^2) -log(x + 12)/log(4)=0}}}


{{{log(x^4)/2log(4) -log(x + 12)/log(4)=0}}}


{{{log(x^4)/2log(4) -2log(x + 12)/2log(4)=0}}}


{{{(log(x^4) -2log(x + 12))/2log(4)=0}}}


{{{(log(x^4) -log((x + 12)^2))/2log(4)=0}}}


{{{log(x^4/(x + 12)^2)/2log(4)=0}}}

will be equal to zero only if


{{{log(x^4/(x + 12)^2)=0}}}..........since {{{log(1)=0}}} (in base {{{10}}})


{{{log(x^4/(x + 12)^2)=log(1)}}}...since log same


{{{x^4/(x + 12)^2=1}}}


{{{x^4=1*(x + 12)^2}}}

{{{x^4=x^2 + 24x+144}}}

{{{x^4-x^2 - 24x-144=0}}}.....factor

{{{(x - 4) (x + 3) (x^2 + x + 12) = 0}}}

solutions:

if {{{(x - 4) = 0}}}=>{{{x=4}}}
if {{{ (x + 3)  = 0}}}=>{{{x=-3}}}=> disregard negative solution for log

if {{{  (x^2 + x + 12) = 0}}}..use quadratic formula

{{{x = (-1 +- sqrt( 1^2-4*1*12 ))/(2*1) }}} 

{{{x = (-1 +- sqrt( 1-48 ))/2 }}} 

{{{x = (-1 +- sqrt( -47 ))/2 }}} => disregard two complex solutions


answer: {{{x=4}}}