Mouse population with halving

Problem

After four cats arrive, the number of mice on a farm (equal to $200$ on day $x=0$) halves every two days. a) Find the function $y=a\cdot b^{x}$ describing the number of mice as a function of time $x$ (in days) and describe its properties (injectivity, monotonicity, concavity). b) Find after how many days the population drops below one mouse.

Solution

a) On day $x=0$ there are $200$ mice, so $a=200$. Since the population halves every $2$ days, $b^{2}=\tfrac12$, hence $b=\left(\tfrac12\right)^{1/2}=\dfrac{1}{\sqrt2}\approx 0{,}7071$. Therefore $y=200\left(\tfrac12\right)^{x/2}=200\cdot 2^{-x/2}.$ Since $0<b<1$ the function is decreasing, hence injective; its range is $(0,+\infty)$ (not surjective onto $\mathbb{R}$) and it is convex. b) We require $200\cdot 2^{-x/2}<1$: $2^{-x/2}<\frac{1}{200}\ \Rightarrow\ -\frac{x}{2}<\log_2\frac{1}{200}\ \Rightarrow\ x>2\log_2 200=\frac{2\ln 200}{\ln 2}\approx 15{,}29.$ The population falls below one from the $16^{\text{th}}$ day on. $\boxed{\,y=200\cdot 2^{-x/2}\,;\qquad x>2\log_2 200\approx 15{,}29\ \text{days}\,}$