How can I determine a formula for an exponential ratio? I am not very experienced in mathematical notation, so please excuse some terminology misuse or formatting shortcomings.
I have a project in which a value needs to increase from a set minimum to a set maximum in a set number of seconds. It is easy to calculate the linear value based on ratios.
Let $v$ = desired value, $n$ = minimum limit, $x$ = maximum limit, $t$ = elapsed time, and $t_x$ = allocated time:
$$v = \frac{t}{t_x}(x-n) + n.$$
Thus if my values are:
$$n = 5, x = 90, t_x = 1800 \text{ (half hour)}$$
For elapsed time of $5$ minutes ($600$ s):
$$v = \frac{600}{1800} (90-5) + 5 = 33.3.$$
The problem is I want to change this linear growth to exponential growth, and I'm not sure how to alter the formula.
So instead of $33.3$ at $5$ minutes, I would rather have $13$ for example. (Slow initial change, rapid later change.)
How can I insert an exponential growth factor into my equation and honor the minimum and maximum values allowed?
 A: I will change notation slightly.  Our initial smallest value is $a$, and our largest value, after say $k$ seconds, is $b$. So every second our amount gets multiplied by $(b/a)^{1/k}$, the $k$-th root of $b/a$.  At elapsed time $t$ seconds, where $0 \le t \le k$, the value is 
$$a \left(\frac{b}{a}\right)^\frac{t}{k}.$$
This is what would happen if we have an initial population $a$ of bacteria, growing under ideal conditions, and ending up with population $b$ after $k$ seconds. The formula above gives the population at time $t$, where $0 \le t \le k$.
It is also what happens if we have an initial amount of money $a$, which under continuous compounding grows to $b$ in time $k$. 
Remark: The quantity $Q$ grows exponentially if and only if the quantity $\log Q$ grows linearly. So alternately, you could translate your knowledge about linear growth to a formula about exponential growth.
A: Let your model be $v(t) = v_0 e^{\alpha t}$, where $v_0$ and $\alpha$ are constants to be determined.
From your data, you want $v(0) = n$, and $v(t_x) = x$.
This immediately gives $v_0 = n$, and then we have $v(t_x) = n e^{\alpha t_x}$, 
from which we get $\alpha = \frac{1}{t_x} \ln \frac{v(t_x)}{n} = \frac{1}{t_x} \ln \frac{x}{n}$.
So the model is
$$ v(t) = n e ^ { \frac{t}{t_x} \ln \frac {x}{n}}.$$
A: I see several sorts of ways to proceed. So let's suppose we had minimum $5$, max $90$, and $1800$ seconds to get there. Then we might have an exponential of the form $Pe^{rt}$ or perhaps $Pe^{rt} + c$. We might choose $f(t) = e^{rt} + 4$, so that $f(0) = 5$ (the minimum). We ask ourselves what $r$ would make it so that $f(1800) = 90$, the max?
Then we'd look at $e^{r1800} = 90 - 4 = 86$, or $\ln 86/1800 = r$ (that's really small). This would give the start of the exponential, growing slowly and then moreso.
But every choice was arbitrary, so let's look at a few other things you might do. Perhaps you want to change the rate of growth. You might do $5e^{rt}$, finding the correct $r$ again. Or you might do something like $Pe^{r(t - 10)}$, shifting along the exponential curve to places that change more slowly, etc.
