Question 285871
log3(3x)=log3x+log3(4-x)



log3(3x)=log3[x(4-x)]


3x=x(4-x)


3x=4x-x^2


x^2-4x+3x=0


x^2-x=0


x(x-1)=0


x=0 or x-1=0


x=0 or x=1


Answer: only solution is x=1