## Exponential Growth

This post about exponential growth is part of a series of posts to help you prepare for the Advanced Algebra and Functions part of the Accuplacer test.

## Question

Last month, a rumor started on the Internet. Today 1000 people believe it. The number of rumor believers doubles every 2 days. Write an equation for the number of rumor believers, n, x days after today.

## Solution

The number of believers increases in proportion to the number of believers already in existence. That is exponential growth.

How can we set up a formula?

Every multiplication by 2 means taking 2 to one more power. A multiplication by 2 happens every 2 days, so 2 is taken to the power of half the number of days that have passed : x/2. That means 2 needs to the power of x/2. Then multiply  by the baseline number, 1000:

$\fn_jvn {\color{Blue} n=1000 \cdot 2^{x/2}}$

To check, try a couple of points where you can calculate both with and without the formula. Choose small numbers to make it easy. Choose even numbers to avoid fractions in formulas. Let’s see what happens 2 days from today, when $\inline \fn_jvn x=2$, and 4 days from today, when $\inline \fn_jvn x=4$.

After 2 days, the number of believers should have doubled from today’s number. So at $\inline \fn_jvn x=2$, n=2000.

Per the formula, at $\inline \fn_jvn x=2$, $\inline \fn_jvn n=1000 \cdot 2^{2/2}=1000 \cdot 2 = 2000$.

After 4 days, the number of believers should have doubled from the number 2 days from today, which itself is double today’s number. So after 4 days, the number of believers should have quadrupled and at $\inline \fn_jvn x=4$, $\inline \fn_jvn n=4000$.

Per the formula, at $\inline \fn_jvn x=4$, $\inline \fn_jvn n=1000 \cdot 2^{4/2}=1000 \cdot 2^2=1000 \cdot 4 = 4000$.

So the equation $\inline \fn_jvn n=1000\cdot 2^{x/2}$ checks out.

Generally, the format of exponential growth equations looks something like  $\inline \fn_jvn N(t)=N_0b^{kt}$, with the letters changing depending on the application. Here, N(t) represents the number of something present at time t, N0 represents the amount present at time t=0, b is the exponential base (2 in the rumor example, where it stands for doubling), and k is a constant that depends on the rate of growth. Another example: For an investment with interest compounded continuously, the formula is $\inline \fn_jvn A(t)=A_0e^{rt}$, where A is the amount present at time t, A0 is the amount present at time t=0, and r is the rate of interest, expressed as a decimal.

As you can see, you can work some problems of exponential growth, like this one, without using the formula.

## Further Thought

Note that the problem tells you there are 1000 rumor-believers today. Since the number doubles every 2 days, there must have been 500 believers the day before yesterday, and 250 two days before that. When did all this start? One way to find out is to keep dividing by 2 until you get to 1, representing that one person who started it all.

 Day Number of Rumor Believers today 1000 2 days ago 1000/2=500 4 days ago 500/2=250 6 days ago 250/2=125 8 days ago 125/2, about 63 10 days ago 63/2, about 32 12 days ago 32/2 = 16 14 days ago 16/2=8 16 days ago 8/2=4 18 days ago 4/2=2 20 days ago 2/2=1

There was some rounding, but it looks like it took about 20 days to get from one person to 1000.

To confirm that result, start with 1 rumor believer and keep doubling that number until you get to 1000. How many doublings does it take? 10 doublings get you to 1024, close enough. And with a doubling every 2 days, 10 doublings take 20 days. For a related problem, see Exponential Decay.

Interesting factoid: To work with an equation for exponential growth or decay, you don’t have to know the starting point; you just need to know data at some known point.

Related: How to Solve an Exponential Equation.

The Accuplacer sample problem on which this one is based is multiple choice. It is #8 in the sample questions for the Accuplacer Advanced Algebra and Functions test.