set y = f(x) and the equation becomes y = (1/4)^x.
replace y with x and x with y and the equation becomes x = (1/4)^y.
now you want to solve for y.
x = (1/4)^y if and only if log(1/4)(x) = y
your equation becomes y = log(1/4)(x)
set y = f-1(x) and the equation becomes f-1(x) = log(1/4)(x)
the graph of both equations is shown below.
since they are inverse equations, they are reflections about the equation of y = x.
they are inverse equations if (x,y) in f(x) is equal to (y,x) in f-1(x).
the intersection of the both graphs with a line perpendicular to the line y = x shows this to be true.
the red curve is the normal equation and the blue curve is the inverse equation.
the reflection of the point (-1.448,7.448) in the normal equation is the point (7.448,-1.448) in the inverse equation, confirming that they are inverse equations.
in general, if you start with y = bx, and you want to find the inverse equation, then replace x with y and y with x to get the equation x = by
this is true if and only if y = logb(x)
the inverse equation to y = bx is therefore y = logb(x).
here's a nice reference on logarithmix and exponential functions you might find useful.
http://www.themathpage.com/aprecalc/logarithmic-exponential-functions.htm