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
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
(Note: k ≈ 0.896, but it is best to keep it as ln(6)/2 until we do our final calculations.)
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),
Find the pressure 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
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!)
