Exponential question

Question

I'm not quite sure how they made the transition from the first step to the second. Is there a specific property of exponentials that they have used? Any help would be much appreciated.

Answer 1

Your integral is $$-\int\limits_0^t 1.5\times \mu_{x+s}\,\mathrm ds = -1.5\int\limits_0^t \mu_{x+s}\,\mathrm ds.$$ Now since $\exp[u] = e^u$ for all constants $u$, then we have that $$\exp\,\left[-\int\limits_0^t 1.5\times \mu_{x+s}\,\mathrm ds\right] = \exp\,\left[-1.5\int\limits_0^t \mu_{x+s}\,\mathrm ds\right],\tag1$$ and since $e^{uv} = (e^u)^v = (e^v)^u$ then we obtain that $$(1) = \exp\,\left[-\int\limits_0^t \mu_{x+s}\,\mathrm ds\right]^{1.5}.\tag2$$ By letting the integral equal $v$, do not confuse this as $e^{v^u}$. As a matter of fact, $$e^{v^u} = \exp\,\left[\left(-\int\limits_0^t \mu_{x + s}\,\mathrm ds\right)^{1.5}\right] \neq (2).$$ Instead, $(2) = (e^v)^u$. Now all that remains is to solve the integral.

Answer 2

$$\left (e^a\right )^b=e^{ab}$$

Answer 3

The property of exponentials mentioned on the answers can be applied in this notation because, by convection, $\exp(a)^b$ is interpreted as $(e^a)^b$ and not $ e^{(a^b)}$ so: $$ \exp(a)^b=[\exp (a)]^b \ne \exp[a^b] $$