Doubling time

Question

A population doubles in size every 15 years. Assuming exponential growth, find the annual growth rate. 

Answer 1

"A population doubles in size every 15 years. Assuming exponential growth, find the annual growth rate."

Since populations grow continuously, we should use a continuous exponential growth model. In fact, this is the model defined as Exponential Growth (see http://mathworld.wolfram.com/ExponentialGrowth.html)

A = P e^(r t)

If the initial population is 1 unit, then in 15 years the population will be 2 units.

2 = 1 e^(15 r)

Take the Natural Logarithm of both sides:

ln(2) = 15 r

r = ln(2)/15 per year, exactly, or

r ≈ 4.620981203733% per year, approximately (no matter how many decimals you get from a calculator the answer is always approximate).

Since no mention was made of how to round, I give both the exact answer and the most decimals I can get out of my calculator for an approximate answer.

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To use the Compound Interest Formula instead of the Exponential Growth Formula in this problem is wrong. Here’s why:

The Compound Interest Formula states A = P(1 + r/n)^(nt) where the interest is calculated ONLY at integer multiples of t/n; i.e., every 1/n time units (e.g., years).

So A(t) is actually the step function,

A(t) = P(1 + r/n)^(floor(nt)).

As a visualization example, this GeoGebra graph:
http://www.wyzant.com/resources/files/264949/continuous_exponential_vs_discrete_compound_growth

shows:

S(t) = (1 + 0.05/2)^(floor(2 t)),

C(t) = e^(0.05 t), and their difference

D(t) = C(t) – S(t).

You can clearly see that for t > 0, D > 0 and increasing on average.

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Why does this level of detail matter, anyway?

Well, if a bridge an engineer (you?) designed collapses due to a design flaw, “I’m sorry!” will not be a strong defense in the ensuing multibillion dollar law suits brought by the survivors of the victims.

Answer 2

 P ( 1+ r) ^15 = 2P

 

    15 log ( 1+r) = log 2

 

       15 = log 2 / log  1+r)

 

      log ( 1+ r) 15= log2

 

        log ( 1 +r) = (log 2) /15

 

                         = 0.02

 

                1+r = 10 ^0.02= 1.047

 

              r = 0.047   4.7%

Answer 3

Hi Asma,

 

Sometimes the trouble people have with these is not knowing how to set up "doubling" into the equation, but it's actually quite simple.  You could literally use any numbers you like, as long as your future value in 15 years is twice what you start with.  And 1 for today and 2 for 15 years from now works just as well as anything, and is easier cause 1 is always an easy number. :-)

 

I'm going to assume you must use the equation to solve for this.

 

Hence:

 

future amt = present amt (1 + r)ⁿ

 

So 2 = 1 (1 + r)¹⁵  where you are solving for r.

 

 

You also may be having trouble with the solving part of that.  I'll do an example using tripling in 20 years:

 

3 = 1 (1 + r)²⁰

 

First we can just drop that 1, since 1 times anything is itself.  If it helps to think of it like this, then divide both sides by 1 to get rid of it.  The 1 will be gone from the right and the left will still be 3 anyway.

 

3 = (1 + r)²⁰

 

Now we get to the tricky part.  We need to deal with the exponent.  To eliminate an exponent, you need to take the nth root of both sides, in this case the 20th root.  The 20th root of a 20th exponent will rid you of that exponent, leaving only 1 + r:

 

20√3 = 20 √[(1 + r)²⁰]

 

(You'll have to deal with how to get 20th root of 3 on your particular calculator.)

 

1.05646... = 1 + r

 

Now you can just subtract 1 from both sides and you're left with:

 

.05646 = r

 

Or r = ~5.6%