Question 6480

 I suppose the base is 10,if not similar solution.
  f(x)=log(4x)+log((3+x)/x^2)
 Since numbers inside log must be positive,so
 4x > 0 and (3+x)/x^2.
 Hence, x > 0 and 3+x > 0 (or x > -3).
 Therefore,x > 0 and (0, +oo) is the domain of f.
 
 Let y =log(4x)+log((3+x)/x^2) = log (4x(3+x)/x^2) = log (4(3+x)/x)
 = log (4 + 12/x).
 By def. of log, 4 + 12/x = 10^y
  so 12/x = 10^y - 4, we get x = 12/(10^y - 4).
 Since  x > 0, 10^y - 4 >0 , so 10^y >4 so, y > log 4.
 Exchange x,y , we obtain the inverse function g(x) = 12/(10^x -4)
 
a.The inverse of f is g(x) = 12/(10^x -4).

b. Domain of the inverse function (log 4, +oo)

c. Check f(g(x)) = log (4 + 12/[12/(10^x -4)]) 
  =  log (4 + 10^x -4 ) = log (10^x) = x.
 and g(f(x)) = 12/(10^[log (4 + 12/x)] -4)
     = 12 / (4 + 12/x - 4)
     = x.
 This shows g is the inverse function of f.

 Kenny