Question 122347
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Your equation is:  {{{log(3,(x+2))-log(3,(x-3))=log(3,6)}}}


First apply the difference of logs rule:
{{{log(b,a)-log(b,c)=log(b,(a/c))}}}


{{{log(3,(x+2))-log(3,(x-3))=log(3,6)}}}
{{{log(3,((x+2)/(x-3)))=log(3,6)}}}


Now, if {{{log(b,a)=log(b,c)}}} then {{{a=c}}}, so


{{{log(3,((x+2)/(x-3)))=log(3,6)}}} => {{{(x+2)/(x-3)=6}}}


{{{(x+2)/(x-3)=6}}}
{{{x+2=6(x-3)}}}
{{{x+2=6x-18}}}
{{{x-6x=-18-2}}}
{{{-5x=-20}}}
{{{x=4}}}


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Check the answer
{{{log(3,(x+2))-log(3,(x-3))=log(3,6)}}}
{{{log(3,(4+2))-log(3,(4-3))=log(3,6)}}}
{{{log(3,6)-log(3,1)=log(3,6)}}}

But {{{log(b,1)=0}}}, so
{{{log(3,6)=log(3,6)}}}, Answer checks.



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