Use L31 to derive L21.

The given function which has to prove is$L\left(\frac{{\in }^{-1}-e^{-k}}{d}\right)=\frac{In\left(p+b\right)}{In\left(p+a\right)}$

A transformation of a function f(x) into the function g(t) that is useful especially in reducing the solution of an ordinary linear differential equation with constant coefficients to the solution of a polynomial equation.

The inverse Laplace transform of a function F(s) is the piecewise-continuous and exponentially-restricted real function f(t)

The properties which are used:

$$\begin{gathered} L\left(31\right):L\left\{\frac{R\left(\alpha \right)}{i}\right\}={\int }_{p}G\left(u\right)du \\ L\left(2\right);L\left(e^{-an}\right)=\frac{1}{\left(p+s\right)} \end{gathered}$$

From, property (L2) of Laplace transforms,

$L\left(e^{-at}\right)=\frac{1}{\left(p\mathrm{l}a\right)}.......\left(1\right)$

So,

$L\left(e^{-bt}\right)=\frac{1}{\left(p+b\right)}$

Subtract equation (2) from (1).

$$\begin{gathered} L\left(e^{-w}-e^{-br}\right)=\frac{1}{\left(p+w\right)}-\frac{1}{\left(p+b\right)} \\ L\left(e^{ai}-e^{uc}\right)={\int }_p^n\left[\frac{1}{\left(u+a\right)}-\frac{1}{\left(u+b\right)}\right]du \\ L\left(\frac{e^{-e_{-a}-u}}{t}\right)={\left[\mathrm{In}\left(\mathrm{u}+\mathrm{a}\right)-\mathrm{In}\left(\mathrm{u}+\mathrm{b}\right)\right]}_{\mathrm{p}}^{\mathrm{p}} \\ =\left[\mathrm{In}\left(\mathrm{u}+\mathrm{a}\right)-\mathrm{In}\left(\mathrm{u}+\mathrm{b}\right)\right] \\ ={\left[\frac{\mathrm{n}\left(\mathrm{u}+\mathrm{b}\right)}{\mathrm{In}\left(\mathrm{u}+\mathrm{a}\right)}\right]}_{\mathrm{F}}^{\mathrm{F}} \\ ={\left[\frac{\mathrm{In}\left(1+{\displaystyle \frac{\mathrm{k}}{\mathrm{a}}}\right)}{\mathrm{In}\left(1+\frac{\mathrm{k}}{\mathrm{w}}\right)}-\frac{\mathrm{In}\left(\mathrm{a}+\mathrm{b}\right)}{\mathrm{In}\left(\mathrm{a}+\mathrm{a}\right)}\right]}_{\mathrm{p}}^{=} \\ =\left|\mathrm{In}\left(\mathrm{p}+\mathrm{b}\right)-\mathrm{In}\left(\mathrm{p}+\mathrm{a}\right)\right| \\ \mathrm{L}\left(\frac{{\mathrm{e}}^{-\mathrm{a}-{\mathrm{e}}^{-\mathrm{it}}}}{\mathrm{t}}\right)=\frac{\mathrm{In}\left(p\mathit{+}b\right)}{\mathrm{In}\left(p\mathit{+}a\right)} \end{gathered}$$

Hence, $\mathrm{L}\left(\frac{{\mathrm{e}}^{-e^{-b}}}{\mathrm{t}}\right)=\frac{\mathrm{In}\left(p\mathit{+}b\right)}{\mathrm{In}\left(p\mathit{+}a\right)}$.