# 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?

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)?

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

So:

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!)