The time shifting property states that:
$$ \mathcal{L}(u(t-t_0)f(t-t_0))=e^{-st_0}F(s) $$
Now I am given a question which is:
$$ \mathcal{L}(e^{-4t}u(t-3)) $$
Can I do like I have done here? That the time shift is of 3 seconds, so using the property I get final answer as:
$$ \frac{e^{-3s}}{s+4} $$
Edit: if we are asked to find unilateral Laplace Transform then and only then do I get the above answer?
It's wrong! Go step by step.
Step $1$: Find frequency shift due to $e^{-4t}$ using:
If $\mathcal{L}(f(t))=F(s)$, then $\mathcal{L}(e^{-4t}f(t))=F(s+4)$. (which is the answer)
Step $2$: Find $\mathcal{L}(f(t))$ by the property of time shift as follows: $$\mathcal{L}(u(t-3).1)=\frac{e^{-3s}}{s}$$ where $f(t)=u(t-3).1$
So that finally $$F(s+4)=\frac{e^{-3(s+4)}}{s+4}$$