Question 134542
You either meant: {{{2*log(3,(x+4))=log(3,(9+2))}}} or {{{2*log(3,(x+4))=(log(3,9))+2}}}


Let's assume you meant the first one:

{{{n*log(b,x)=log(b,x^n)}}}, so:


{{{log(3,(x+4)^2)=log(3,11)}}}


=> {{{(x+4)^2=11}}}


{{{x+4=sqrt(11)}}} or {{{x+4=-sqrt(11)}}}


and finally, {{{x=-4+-sqrt(11)}}}


However, since {{{9<11<16}}}, {{{3<sqrt(11)<4}}}.  That means that {{{-4+sqrt(11)>-1}}} and {{{-4-sqrt(11)<-7}}}.


The domain of {{{log(b,x)}}} is {{{0<x<infinity}}}, so the domain of {{{log(3,(x+4))}}} must be {{{-4<x<infinity}}}


Therefore, {{{-4-sqrt(11)}}} must be excluded since it represents a value that would make {{{x+4}}} be outside of the domain of {{{log(3,(x+4))}}}.  


{{{-4+sqrt(11)+4}}} is in the domain of {{{log(3,(x+4))}}}, therefore {{{x=-4+sqrt(11)}}} completely describes the solution set for the given equation.


On the other hand, if you really meant:


{{{2*log(3,(x+4))=(log(3,9))+2}}}


Then begin with the same rule for logarithms used above, namely: {{{n*log(b,x)=log(b,x^n)}}} to write:


{{{log(3,(x+4)^2)=(log(3,9))+2}}}


Since {{{log(b,x)=y}}} if and only if {{{b^y=x}}}, we can say {{{log(3,9)=y}}} if and only if {{{3^y=9}}} and therefore {{{log(3,9)=y=2}}}, hence:


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


Again using {{{log(b,x)=y}}} if and only if {{{b^y=x}}}, we can write:


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


Taking the square root of both sides:


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


And
{{{x=5}}} or {{{x=-13}}}.  Using the same logic as above, exclude any root less than -4.  {{{x=5}}} is the single element of the solution set.