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) = e^{x}
 At 1 year old it is: e^{1 } = 2.7 mm high ... really tiny!
 At 5 years it is: e^{5 } = 148 mm high ... as high as a cup
 At 10 years: e^{10 } = 22 m high ... as tall as a building
 At 15 years: e^{15 } = 3.3 km high ... 10 times the height of the Eiffel Tower
 At 20 years: e^{20 } = 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_{} × e^{kt}
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

Start with the formula:
y(t) = a_{} × e^{kt}
We know a_{}=3 mice, t=2 months, and right now y(2)=18 mice:
18 = 3_{} × e^{2k}
Now some algebra to solve for k:
Divide both sides by 3:  6 = _{} e^{2k}  
Take the natural logarithm of both sides:  ln(6) = _{} ln(e^{2k})  
ln(e^{x})=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),
 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_{} × e^{kt}
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_{} e^{k×1000}
Now some algebra to solve for k:
Divide both sides by 1013:  0.88 = _{} e^{1000k}  
Take the natural logarithm of both sides:  ln(0.88) = _{} ln(e^{1000k})  
ln(e^{x})=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!)