Is This a Mistake?: Regarding the property $e^x e^y = e^{x+y}$

Question

My textbook has the following:

$$e^{\lambda(s-1)} e^{\mu (s-1)} = e^{(\lambda+\mu)s-1}$$

This is not in the errata but I felt it should be,

$$e^{\lambda(s-1)} e^{\mu (s-1)} = e^{(\lambda+\mu)(s-1)}$$

My apologies if I made a trivial or silly mistake. I'm checking my work and I simplified to the second equation but if it's the first one would you mind explaining how that is? Thank you.

Again, this does not appear in the errata provided.

Answer 1

Yes of course since

$$a^x\cdot a^y=a^{x+y}$$

the following is right

$$e^{\lambda(s-1)} e^{\mu (s-1)} =e^{\lambda(s-1)+\mu (s-1)} = e^{(\lambda+\mu)(s-1)}$$

Answer 2

$$e^{\lambda(s-1)} e^{\mu (s-1)} = e^{(\lambda+\mu)s-1}$$ $$e^{(\lambda+\mu)s} e^{-(\mu +\lambda)} = e^{(\lambda+\mu)s}e^{-1}$$ $$e^{-(\mu +\lambda)} = e^{-1}$$ Only true for $$\mu = 1-\lambda$$ But then he would have expressed this like this $$e^{(\lambda+\mu)s} e^{-(\mu +\lambda)} = e^{s-1}$$ When this is true for all $\lambda,\mu$ $$e^{\lambda(s-1)} e^{\mu (s-1)} = e^{(\lambda+\mu)(s-1)}$$ Depends on the context because first equality may be true in some contexte while the second equality is always true.