Let’s take the exponential growth function to be $ y= A_0e^{kt}$ and exponential decay function to be $ y= A_0e^{-kt} $ .
I’m just told that to find out How fast the process is, you have to double the exponential growth function and Half-time the exponential decay function. Why is this so? I never seem to understand it. Is there a way to prove it? Or is my tutor right by just telling me the fact and nothing else?
On
You do not have to. If you know $f(t_0)$ and f$(t_1)$ for any distinct points $t_0$ and $t_1$, you can find that $$ \frac{f(t_1)}{f(t_0)}=\frac{Ae^{kt_1}}{Ae^{kt_0}}=e^{k(t_1-t_0)}$$ and therefore $$\tag1 k=\frac1{t_1-t_0}\ln\frac{f(t_1)}{f(t_0)}.$$ It is simply a convenient convention to pick $t_1$ such that $f(t_1)$ is $2f(t_0)$ or $\frac12f(t_0)$ (depending on growth or decay). A more natural choice (somewhat literally, as we are taking natural logarithms) would be to find $t_1$ such that the value is multiplied or divided by $e$ instead of $2$. At any rate, both would require something like permanent observation - and certainly nobody has observed the decay process of Uranium long enough (billions$^ 1$ of years) for either of these methods. Instead, one makes two (or more) observations at suitable values of $t$ and "aims" for a ratio of approximately $2$ (or $\frac12$), even though theoretically any ratio would work.
But still there is a practical aspect as to why we prefer the ratio to be approximately $2$: For example, if we have a quickly decaying substance and measure (with precision clocks but a balance wit $0.1$ gram precision) that an initial mass of $100\pm0.1$ grams decays into $0.2\pm 0.1$ grams in $10$ seconds, we can determine $k$ only with low precision from $(1)$, something like $-0.691\le k\le -0.581$ (by plugging in a high value for $f(t_0)$ and a low value for $f(t_1)$ or vice versa; numbers here and in what follows are rounded to three decimals). Similarly, if we measure $94.2\pm0.1$ grams after $0.1$ seconds, we find something like $-0.618<k<-0.577$. But if we measure $54.9\pm.1$ grams after $1$ second, we obtain $-0.602<k<-0.597$. Apparently, the last case gives us the best precision for the final result, so the "sweet spot" must somewhat balance between measurement errors compared to $f(t_1)$ as well as compared to $|f(t_1)-f(t_0)|$. If $f(t_0)$, $f(t_1)$, and $|f(t_1)-f(t_0)|$ are all of the same order of magnitude, we are good. It happens precisely at a factor of $2$ or $\frac12$ that two of these three numbers are equal and we have the minimal "spread" between the sizes of these three. (By the way, all example numbers above were produced by rounding $100\cdot e^{0.6t}$)
$^1$ US billions only, but still ...
Let's take the exponential growth function $f(t)=A_0e^{kt}$. It is simple to find $A_0$ - this is just the value of $f$ at time $0$. But how can we find the value of $k$ ?
Suppose $f(t)$ has value $y_0$ at time $t_0$ and value $2y_0$ at time $t_1$. Then
$2y_0=2A_0e^{kt_0}=A_0e^{kt_1}$
$\Rightarrow 2e^{kt_0}=e^{kt_1}$
$\Rightarrow \ln(2) + kt_0=kt_1$
$\Rightarrow k=\frac{\ln(2)}{t_1-t_0}$
The time interval $t_1-t_0$ during which $f(t)$ doubles is called the "doubling time". Notice that it does not depend on when it is measured. Once you know the doubling time you can find the growth constant $k$.
You can do the same calculation with exponential decay except you measure the time taken for $f(t)$ to halve, not double, which is called the "half life".