Question 928299
1. solve it

{{{log(2,(x-3))=3-log(2(x-1))}}} 


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

{{{log(2,((x-3)(x-1)))=3}}} ...in base {{{10}}} 

{{{log(((x-3)(x-1)))/log(2)=3}}}

{{{log(((x-3)(x-1)))=3log(2)}}}

{{{log(((x-3)(x-1)))=log(2^3)}}}

{{{log(((x-3)(x-1)))=log(8)}}} ...if log. same, then

{{{(x-3)(x-1)=8}}} is same too; so, solve it for {{{x}}}

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

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

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

{{{x^2+x-5x-5=0}}}...group

{{{(x^2+x)-(5x+5)=0}}}

{{{x(x+1)-5(x+1)=0}}}

{{{(x-5)(x+1)=0}}}

solutions:

if {{{x-5=0}}}=> {{{highlight(x=5)}}}...your answer

if {{{x+1=0}}}=> {{{x=-1}}}...reject negative value for log


2. express it as a single logarithm

{{{6log(7,(x^2-1))-5log(7,(x-1))}}} 


{{{log(7,(x^2-1)^6)-log(7,(x-1)^5)}}}


{{{log(7,((x^2-1)^6/(x-1)^5))}}}....since {{{(x^2-1)^6=((x-1)(x+1))^6=(x-1)^6(x+1)^6}}}, we will have


{{{log(7,(((x-1)^6(x+1)^6)/(x-1)^5))))}}}


{{{log(7,(((x-1)^cross(6)^1(x+1)^6)/cross((x-1)^5)1))}}}


{{{log(7,((x-1)(x+1)^6))}}}