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 $x \in R$.

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 $g_{-k}(x)= \log_b(x-k)$.

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 $.

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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 $f$.