Exponential Growth and Decay

Exponential growth can be amazing!


Let us say we have this special tree. It grows exponentially, following this formula (e is Eulers number):

Height (in mm) = ex

  • At 1 year old it is: e1 = 2.7 mm high ... really tiny!
  • At 5 years it is: e5 = 148 mm high ... as high as a cup
  • At 10 years: e10 = 22 m high ... as tall as a building
  • At 15 years: e15 = 3.3 km high ... 10 times the height of the Eiffel Tower
  • At 20 years: e20 = 485 km high ... up into space!

No tree could ever grow that tall!
So when people say "it grows exponentially" ... just think what that means.

Growth and Decay

But sometimes things can grow (or the opposite: decay) exponentially, at least for a while.

So we have a generally useful formula:

y(t) = a × ekt

Where y(t) = value at time "t"
a = value at the start
k = rate of growth (when >0) or decay (when <0)
t = time


Example: 2 months ago you had 3 mice, you now have 18.


Assuming the growth continues like that

  • What is the "k" value?
  • How many mice 2 Months from now?
  • How many mice 1 Year from now?


Start with the formula:

y(t) = a × ekt

We know a=3 mice, t=2 months, and right now y(2)=18 mice:

18 = 3 × e2k

Now some algebra to solve for k:

Divide both sides by 3:   6 = e2k
Take the natural logarithm of both sides:   ln(6) = ln(e2k)
ln(ex)=x, so:   ln(6) = 2k
Rearrange:   k = ln(6)/2


Now, we want to know the population in 2 more months (at t=4 months), and in 1 year from now (t=14 months):


y(4) = 3 e(ln(6)/2)×4 = 108

y(14) = 3 e(ln(6)/2)×14 = 839,808


That's a lot of mice! I hope you will be feeding them properly.

Exponential Decay

Some things "decay" (get smaller) exponentially.

Example: Atmospheric pressure (the pressure of air around you) decreases as you go higher.

It decreases about 12% for every 1000 m: an exponential decay.

The pressure at sea level is about 1013 hPa (depending on weather).

  • Write the formula (with its "k" value),
  • What would the pressure be on the roof of the Empire State Building (381 m),
  • and at the top of Mount Everest (8848 m)?


Start with the formula:

y(t) = a × ekt

We know

  • a (the pressure at sea level) = 1013 hPa
  • t is in meters (distance, not time, but the formula still works)
  • y(1000) is a 12% reduction on 1013 hPa = 891.44 hPa


891.44 = 1013 ek×1000

Now some algebra to solve for k:

Divide both sides by 1013:   0.88 = e1000k
Take the natural logarithm of both sides:   ln(0.88) = ln(e1000k)
ln(ex)=x, so:   ln(0.88) = 1000k
Rearrange:   k = ln(0.88)/1000


Now we know "k" we can write:

y(t) = 1013 e(ln(0.88)/1000)×t


And finally we can calculate the pressure at 381 m, and at 8848 m:


y(381) = 1013 e(ln(0.88)/1000)×381 = 965 hPa

y(8848) = 1013 e(ln(0.88)/1000)×8848 = 327 hPa


(In fact pressures at Mount Everest are around 337 hPa ... not bad!)