What is the laplace transform s=4 ? h(t)=e−3(t−2)u(t−2)

Step 1
Given: ${\displaystyle h\left(t\right)=e^{-3(t-2)}u(t-2)}$
To find the laplace transform of the function h(t) and what is the ${\displaystyle L\cdot T}$ $s=4$ ?
Step 2
We know that
$L\{f(t-a)\cdot u(t-a)\}=e^{-as}\cdot F(s)$
$\therefore L\{e^{-3(t-2)}u(t-2)\}=e^{-2s}\cdot F(s)$
Let ${\displaystyle f\left(t\right)=e^{-3t}}$
$\Rightarrow L\{f(t)\}=L\left\{e^{-3t}\right\}$
${\displaystyle =\frac{1}{s+3}=F\left(s\right)}$
$\therefore L\{e^{-3(t-2)}\cdot u(t-2)\}=e^{-2s}\cdot \frac{1}{s+3}$
Now at $s=4$ , $L\{h(t)\}=\frac{e^{-2\times 4}}{4+3}$
${\displaystyle =\frac{1}{7}{e}^{-8}}$