Q:

plant height $h(t)$ increases at constant rate $k$%. if in $100n$ hours height is doubled, prove $k=\frac{1}{n}\ln2$

plant height $h(t)$ increases at constant rate $k$%. if in $100n$ hours plant height is doubled, prove $k=\frac{1}{n}\ln2$
My Try
I used compound interest theory to calculate this;
$2h=h(1+\frac{k}{100})^{100n}$
But I cannot get the required answer. What is my mistake? Thanks a lot.

A:

If the plant grew in spurts once a year you would calculate
$h(t) = h(1 + .01*k)^t$.
If the plant grew in spurts $m$ times a year you would calculate
$h(t) = h(1+ .01*\frac km)^{mt}$.
But if the plant grew continuously it you would calculate
$h(t) = h\lim_{m\to \infty} (1+ .01*\frac km)^{mt}=h\lim_{m\to \infty}[ (1 + \frac 1m)^m]^{0.01* kt} = he^{0.01*kt}$.
So $2h = h(100n) = h e^{0.01*k*100n} = he^{nk}$
So $2 = e^{nk}$ so
$\ln 2 = nk$ and 
$k = \frac {\ln 2} n$
====
Euler's number is often (not always) defined to be $\lim_{n\to \infty} (1 + \frac 1n)^n$.  This is useful for continuous growth question.  
If something is growing at a constant rate of $r$ (per time unit) and but growing continually then the compounded growth after time $t$ will be $\lim_{n\to \infty} (1+ \frac rn)^{nt} = \lim_{n\to \infty} [(1+\frac 1n)^n]^{rt} = e^{rt}$.
This is based on $\lim_{n\to \infty} (1 + \frac an)^n =\lim_{n\to \infty}(1 + \frac 1n)^{an} = e^a$.
But you asked in a comment why we can move the $a$ for $(1+\frac an)^n$ to the exponent $(1 + \frac 1n)^{an}$?
Well.....
$\lim_{n\to \infty} (1 + \frac an)^n=$
$\lim_{n\to \infty} (1 + \frac 1{\frac na})^{a\cdot \frac na}$.
Let's replace then notation of $\frac na$ with $m$ so
$\lim_{n\to \infty} (1 + \frac 1{\frac na})^{a\cdot \frac na}=$
$\lim_{am \to \infty}(1 + \frac 1m)^{am}$.
Now $am\to \infty$ as $m\to \infty$ so 
$\lim_{am\to \infty}(1 + \frac 1m)^{am} =$
$\lim_{m \to \infty}(1 + \frac 1m)^{am} =$
$\lim_{m\to \infty}[(1 + \frac 1m)^m]^a =$
$e^a$.

