Let \(E(t) = 0.5 \, e^{-t-3} - 4\).

Graph of \(y = E(t) = 0.5 \, e^{-t-3} - 4\).

We have a function called \(E(t)\). Its pairs look like \((a, E(a))\). The inverse of this function will have pairs of the form \((E(a), a)\). The domain and range just switch roles.

In the exponential formula and graph, \(E\) is representing the range and \(t\) the domain. They just switch roles for the inverse. Same equation. Just a different interpretation.

Now, \(t\) is the function name and \(E\) is the variable. We can just solve this for \(t(E)\).

\begin{align*} E & = 0.5 \, e^{-t(E)-3} - 4 \\ \answer {E + 4} & = 0.5 \, e^{-t(E)-3} \\ 2(E + 4) & = e^{-t(E)-3} \\ \ln (2(E+4)) & = \answer {-t(E)-3} \\ -(\ln (2(E+4))) & = t(E)+3 \\ \answer {-(\ln (2(E+4))) - 3} & = t(E) \end{align*}

Graphs of \(y = E(t) = 0.5 \, e^{-t-3} - 4\) and \(z = L(x) = -ln(2(x+4)) - 3\).

Since, these are inverses of each other, the pairs of each function are just reversed and the graphs are mirror images of each other over the graph of the identity function: \(\{ (r,r) \, | \, r \in \mathbb {R} \}\).

• The domain of \(E(t)\) is \(\mathbb {R}\). The range of \(L(x)\) is \(\mathbb {R}\).
• The range of \(E(t)\) is \((-4, \infty )\). The domain of \(L(x)\) is \((-4, \infty )\).

• The graph of \(E(t)\) has a horizontal asymptote at \(-4\).
• The graph of \(L(x)\) has a vertical asymptote at \(-4\).

\(L(x) = \ln (x+5) - 1\) is a one-to-one function, as its graph illustrates.

Therefore, it must have an inverse function, \(x(L)\).

We can just solve for \(x\).

\begin{align*} L & = \ln (x+5) - 1 \\ \answer {L + 1} & = \ln (x+5) \\ \answer {e^{L+1}} & = x+5 \\ e^{L+1} - 5 & = x \\ e^{L+1} - 5 & = x(L) \end{align*}