Rates of growth, and differential equations - What's the rate of doubling?

Question

Problem

Let $y(x) = e^{kx}$.

Moreover, let $y_1 =(1.01)^x$ and $y_2 = (1.02)^x$.

Show that $y_1$ can be written as $y_1 = e^{k_1 x}$ and ditto for $y_2, k_2$.

If $x$ has the unit $\text{years}$, what is the rate of doubling for $y_1$ and $y_2$?

My progress

First of all, $$y_1 = e^{\ln(1.01)x} \Rightarrow k_1 = \ln(1.01)$$ and $$y_2 = e^{\ln(1.02)x} \Rightarrow k_2 = \ln(1.02)$$

Question

I get stuck from here, because I've forgotten about how we find the rate at which an exponential function doubles (or triples etc.) its value.

My instinct is to let $y = 2y_1$, but then the RHS is a function of $x$, and the question calls for a constant value.

What's the basic thing I'm missing here?

Thanks so much in advance!

Answer 1

You want $t_2$ satisfying $2x_0=x_0e^{kt_2}$.

Solving, we get $t_2=\frac{\ln2}k$.

Answer 2

Follow yor instinct.

$y_1 = e^{\ln(1.01)x_1}$

$2y_1=2e^{\ln(1.01)x_1}=e^{\ln(1.01)x_1+\ln(2)}=e^{\ln(1.01)\left(x_1+\frac{\ln(2)}{\ln(1.01)}\right)}$

$y_1(x_1+\frac{\ln(2)}{\ln(1.01)})=2y_1(x_1)$

This is the constant $\frac{\ln(2)}{\ln(1.01)}$. Whenever it is added to the independent variable the dependent variable doubles its previous value.

Answer 3

If you deposit $1$ dollar then the time it takes to have $2$ dollars is found by $$e^{kT} =2$$ which gives you $$T=\frac {\ln2}{k}$$ If you want to triple your money then $$ T=\frac {\ln 3}{k}$$.

Since you have your $k$ values you can find the doubling time.