Inverse functions
Logarithmic functions

Exponential and logarithmic functions
Exponential and logarithmic functions are mutually inverse functions
Inverse functions
The inverse function, usually written f -1, is the function whose domain and the range are respectively the range and domain of a given function f, that is
f -1(x) = y  if and only if   ƒ (y) = x .
Thus, the composition of the inverse function and the given function returns x, which is called the identity function, i.e.,
f -1(ƒ (x)) = x    and    ƒ (f -1(x)) = x.
The inverse of a function undoes the procedure (or function) of the given function.
A pair of inverse functions is in inverse relation.
Example:  If given ƒ (x) = log2 x  then f -1(x)  = 2x  since,

Therefore, to obtain the inverse of a function y = ƒ (x), exchange the variables x and y, i.e., write x = ƒ (y) and solve for y.  Or form the composition ƒ(f -1(x)) = x  and solve for f -1.
Example:  Given y = ƒ (x) = log2 determine  f -1(x).
Solution:  a)  Rewrite  y = ƒ (x) = log2 x  to  x = log2 y  and solve for y, which gives  y = f -1(x) = 2x.
b)  Form ƒ (f -1(x)) = x that is,  log2 (f -1(x)) = x and solve for f -1, which gives f -1(x) = 2x.
The graphs of a pair of inverse functions are symmetrical with respect to the line  yx.
The graph of the logarithmic function  y = logaxa > 0  and  for  a = ey = logex = ln x
The logarithmic function is inverse of the exponential function since its domain and the range are respectively the range and domain of the exponential function, so that

The domain of ƒ (x) = loga x is the set of all positive real numbers.
The range of ƒ (x) = loga x is the set of all real numbers.
If a > 1 then ƒ is an increasing function and if  0 < a < 1 then ƒ is a decreasing function.
The graph of the logarithmic function passes through the point (1, 0). The y-axis is the vertical asymptote to the graph, as shows the above picture.
Translated logarithmic and exponential functions
Example:  Given translated logarithmic function  y - 2 = log3 (x + 1), find its inverse and draw their graphs.
Solution:  Exchange the variables and solve for y, that is  x - 2 = log3 (y + 1)  which gives  y + 1 = 3x - 2.
Note that         y - y0 = loga (x - x0)  represents translated logarithmic function to base a
and          y - y0 = a (x - x0)      represents translated exponential function with base a
where,  x0 and  y0  are the coordinates of translations of the graph in the direction of the coordinate axes.
College algebra contents D