Why Does e^6 Not Appear in the Laplace Transform Solution?

[RadiationX]

I need to find the Laplace transform of the function below.

$$2\,tu \left( t \right) +5\,{e^{-3\,t+6}}u \left( t-2 \right)$$

The answer in the back of the book is the following.
$$2\,{s}^{-2}+5\,{\frac {{e^{-2\,s}}}{s+3}}$$


I understand how the first term was arrived at, but the second term is confusing me.
From the rules of exponents, when you multiply exponents you add them. So I can rewrite the second term like this $$5e^{-3t}e^6*u(t-2)$$

Now, shouldn't the constants of 5 and $$e^6$$ become multipliers now? If this is so why don't they appear like that in the final answer?

Any Ideas why $$e^6$$ does not appear in the final answer?

 

[lippyka]

The key insight to understand why e^6 does not appear in the final answer is that the Laplace transform of a unit step function is 1/s. Since e^6 appears in the exponent of the second term, it will cancel out when you take the Laplace transform. Therefore, the final answer is 2/s^2 + 5e^(-2s)/(s+3).