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 Euler's number):


Height (in mm) = ex

e^x graph



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
Swap sides:2k = ln(6)
Divide by 2:k = ln(6)/2

The step where we used ln(ex)=x is explained more at Exponents and Logarithms.

Note: we can calculate k ≈ 0.896, but it is best to keep it as k = ln(6)/2 until we do our final calculations.


We can now put k = ln(6)/2 into our formula (also a=3):

y(t) = 3 e(ln(6)/2)t

Now let's calclulate the population in 2 more months (at t=4 months):

y(4) = 3 e(ln(6)/2)×4 = 108

And in 1 year from now (t=14 months):

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).

mount everest



Start with the formula:

y(t) = a × ekt

We know


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
Swap sides:1000k = ln(0.88)
Divide by 1000: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 ... good calculations!)