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I am trying to understand exponential growth doing some practice questions. I arrived at a solution but they don't feel exponential enough i.e. the value isn't growing as high as I expected.

Given a population of 90 if it grows at 3% per year what will the population be after 4 years.

My logic was to calculate this for each year and round UP as you cannot have 0.5 a person at the end of the year and it should be biased towards a higher full number?

Year 1:(90 x 3)/100 = 2.7 = 90+2.7=92.7 (93 rounded)

Year 2:(93 x 3)/100 = 2.79 = 93+2.79=95.79 (96 rounded)

Year 3:(96 x 3)/100 = 2.88 = 96+2.88=98.88 (99 rounded)

Year 4:(99 x 3)/100 = 2.97 = 99+2.97=101.97 (102 rounded)

Am I on the right track here? It has only grown by 12 after 4 years.

If I use x(t) = x_0 × (1 + r)^t Exponential Growth Calculator I am left with 101.2 (102 rounded) after 20 years the population is 162.5. The growth doesn't seem like its exponentially growing, or is it just because the growth % is so low?

Thanks in advance!

Ben
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  • By the way, It's usually better to round once at the end and to round toward the nearest integer, unless you have a compelling reason to do otherwise. – Fabio Somenzi Mar 31 '17 at 02:47

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You are on the right track. It's just that with a small base it takes a while for the exponential to show its true colors.

And yet, give it enough time and $1.03^n$ will pass $n^{1000}$. Try $n=10^6$ and compare $(10^6)^{1000}$ to $1.03^{10^6}$. The easier way is to take logarithms base $10$:

$$ \log_{10} (10^6)^{1000} = 6000 < \log_{10}1.03^{10^6} = 10^6 \log_{10} 1.03 = 29558.8 \enspace. $$

The first is a number with 6001 digits. Quite large, admittedly. But the second number has over twenty thousand more digits!

In fact any exponential will eventually pass any polynomial as long as the base is greater than 1.

Fabio Somenzi
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