We start by recalling the key concept: The inverse of a
function.
Now consider that the functions
$exp_b$ and $\log_b$ are inverses by definition.
As a result,
functions that have these functions a components either arithmetically
or as compositions will have inverses using the related inverse. If an
exponential function is used, then the related logarithmic function
will appear in the inverse, and vice versa.
Note: The algebra of exponential
and logarithmic functions can often help simplify the
expressions of inverses.
As as example we consider functions of the form $f(x) = A b^{x-h}+k$.
This exponential function is constructed with
compositions using the
exponential function, $exp_b(x) =b^x$, and simple linear functions. Finding the algebraic expression for the inverse
of $f$ can be interpreted visually by composing the inverses
of $f_{-h}(x)=x-h$, $exp_b(x)=b^x$ , $f_A(x)
=A*x$,and $q_{+k}(x)=x+k$ in the reverse order ("socks
and shoes").
Thus the inverse of $f(x) = A*b^{x-h} +k$ is
$g(x)= (f_{+h}\circ \log_b \circ f_{1/A}\circ
f_{-k})(x) = \log_b { \frac {x-k} A} +h $.
We use mapping diagrams to verify and visualize how the
function $g(x) = \log_b ( \frac {x-k} A) +h = \log_b(x-k) -
\log_b(A) + h $ is the inverse of $f$
when $A \gt 0$ by looking at the composition
functions $(f\circ g)(x) = x$ for $x \ge k$ and $(g \circ f)(x) =
x$ for all $x \in R$.
Example
ELF.INV.1 :
Suppose $f(x) = 2b^{x-1} -3$. Verify that $g(x) = \log_b( \frac
{x+3} 2) + 1$ is the inverse function for $f$ for $x \ge -3$.
Inverses for $f_{A*}(x) =A*b^x$
and $f_{+k}(x)=b^x + k$ are quite simple: $g_{\frac 1 A
*}(x) = \log_b( \frac x A )$ and $g_{-k}(x)= \log_b(x-k)$.
Example
ELF.INV.2 : (i) Suppose $f(x)=Ab^x$. [$A \ne0$] Verify that $g(x)=log_b(\frac xA)$ is the inverse function for $f$.
ii) Suppose $f(x)=b^x+k$. Verify that $g(x)=log_b(x−k)$ is the inverse function for $f$.
Since many exponential functions are constructed using the
natural exponential function, $exp_e(x) =e^x$, and simple linear
functions with
composition, finding the algebraic expression for the inverse
of $f$ in this case can be interpreted visually by composing the inverses
of $f_{-h}(x)=x-h$, $exp(x)=e^x$ , $f_A(x)
=A*x$,and $q_{+k}(x)=x+k$ in the reverse order ("socks
and shoes").
Thus the inverse of $f(x) = A*e^{x-h} +k$ is
$g(x)= (f_{+h}\circ \ln \circ f_{1/A}\circ
f_{-k})(x) = \ln { \frac {x-k} A} +h $.
Example
ELF.INV.3 : Suppose $f(x) = Ae^x +k$. Verify
that $g(x) =\ln(\frac {x-k} A)$ is the inverse function for
$f$.
You can use this next dynamic example
to examine visually $g$ as the inverse of $f$ with the
mapping diagrams and the graphs of $g$ and $f$.
Example
ELF.DINV.0.1 (Not Yet Done) Dynamic Visualization of The Inverse Function
for Logarithmic and Exponential Functions: Graph and Mapping Diagrams