<s>[INST] Book: Actuarial Mathematics for Life Contingent Risks Exercise 2.4 Exercise $2.4$  Let $$S_0(x) =\exp\left({-\left( Ax+ \frac{Bx^2}2 +\frac{C(D^x-1)}{\ln D}\right)}\right)$$
where $A, B, C , D>0$
$(b)$ Derive a formula for $S_x(t)$
I know the equation for $S_x(t)= \dfrac{S_0(x+t)}{S_0(x)}$
Right now I am at: $\dfrac{\exp\left({-\left( Ax+ At+\frac{B(x^2+2xt+t^2)}2 +\frac{C(D^{x+t}-1)}{\ln D}\right)}\right)}{\exp\left({-\left( Ax+ \frac{Bx^2}2 +\frac{C(D^x-1)}{\ln D}\right)}\right)}$
Do the $2$ exp's combine to just be a single exp?  I am unsure of how to cancel things out.  Essentially the basic algebra is tripping me up.  Please Help!
 [/INST] It is better to let $$f(x) = Ax + \frac{Bx^2}{2} + \frac{C(D^x - 1)}{\log D}$$ for fixed constants $A, B, C, D$, then observe $$S_x(t) = \frac{S_0(x+t)}{S_0(x)} = \frac{\exp(-f(x+t))}{\exp(-f(x))} = \exp(-f(x+t)+f(x)).$$  Then $$\begin{align}
f(x) - f(x+t) 
&= \left(Ax - A(x+t)\right) + \left(\frac{Bx^2}{2} - \frac{B(x+t)^2}{2}\right) + \left(\frac{C(D^x - 1)}{\log D} - \frac{C(D^{x+t} - 1)}{\log D}\right) \\
&= -At - \frac{B}{2}\left((x^2 + 2xt + t^2) - x^2\right) - \frac{C}{\log D} \left(D^x D^t - D^x \right) \\
&= -\left( At + \frac{B}{2}t(2x + t) + \frac{C D^x}{\log D}(D^t - 1)\right).
\end{align}$$
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