Question 946643


Solve for {{{a}}} in {{{log(2a)-3log(2)= (1/2)log(a-3)}}}

{{{log(2*a)-3log(2)= (1/2)log(a-3)}}}

{{{log(2a)-log(2^3)= log((a-3)^(1/2))}}}

{{{log(2a)-log(8)= log(sqrt((a-3)))}}}

{{{log((2a/8))= log(sqrt((a-3)))}}}

{{{log((a/4))= log(sqrt((a-3)))}}} ...if log same, then

{{{a/4= sqrt(a-3)}}}...square both sides

{{{(a/4)^2= (sqrt(a-3))^2}}}

{{{a^2/16= (a-3)}}}

{{{a^2= 16(a-3)}}}

{{{a^2= 16a-48}}}

{{{a^2-16a+48=0}}}...factor completely

{{{a^2-4a -12a+48=0}}}...group

{{{(a^2-4a) -(12a-48)=0}}}

{{{a(a-4) -12(a-4)=0}}}

{{{(a-12)(a-4) = 0}}}

solutions:

if {{{(a-12) = 0}}}=> {{{a=12}}}

if {{{(a-4) = 0}}}=> {{{a=4}}}