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Exponential growth yields repeating doublings

Exponential decay yields repeating halvings

The amount of time during which this doubling (or halving) takes place is an important characteristic of the process

We already have some examples:

the amount of bacteria doubles every minute, while the amount of magic pennies doubles every day

If we want to know the value of our exponentially growing quantity at a moment of time, the knowledge of doubling time, call it \( T_{\text{double}} \), is of great help

Assume that we know the initial value (at a moment \(t=0\)) of our quantity of interest

We can use the formula

\(
\text{new value} = \text{initial value} \times 2^{t/T_{\text{double}}}
\)

in order to calculate the new value at a moment of time \(t\).

Exercise 26, p490

The doubling time of a bank account balance is 30 years. By what factor does it grow in 60 years? in 90 years?

Solution 1. Let the initial balance be \(\$100 \).

It becomes \(\$200\) 30 years later.

It becomes \($400\) another 30 years later.

Thus \(\$100 \) has turn into \($400\) during 60 years, and the factor is 4.

Similarly, during 90 years \(\$ 100 \rightarrow \$ 200 \rightarrow \$ 400 \rightarrow \$ 800\), and the factor is 8.

Solution 2. Same answers from the formula: \( \text{new value} = \text{initial value} \times 2^{t/T_{\text{double}}} \)

With \(T_{\text{double}} = 30\), the factor in question is

\( 2^{t/T_{\text{double}} }= 2^{t/30} \), and

when t=60 years, \( 2^{t/T_{\text{double}} }= 2^{t/30}=2^{60/30}=2^2=4 \),

while when t=90, \( 2^{t/T_{\text{double}} }= 2^{t/30}=2^{90/30}=2^3=8 \).

\( T_{\text{double}} \) may not be given.

It is an important characteristic of an exponential growth, and we may want to find it

If \(P\) is the percentage growth rate, then

\(
T_{\text{double}} \approx \frac{70}{P}
\)

assuming that \(P\) is relatively small.

Note that \(P\) here is in __ percents __ per unit of time

Oil consumption is increasing at a rate of 2.2% per year. What is the doubling time? By what factor will the oil consumption increase in a decade?

Solution. By the approximate doubling time formula with P=2.2%,

\(
T_{\text{double}} \approx \frac{70}{P} \approx \frac{70}{2.2} \approx 32 \text{years}
\)

The factor is

\(
2^{t/T_{\text{double}}} = 2^{10/32} \approx 1.24
\)

Oil consumption is increasing at a rate of 2.2% per year. By what factor will the oil consumption increase in a decade?

Alternative solution.
Every year the oil price increases by a factor of

\(1+ 0.022=1.022\).

Over a decade, the price thus increases by a factor of

\(1.022^{10} \approx 1.243\).

With decay everything is similar to growth except everywhere is \( \frac{1}{2} \) instead of \(2\), and \(T_{\text{half}}\) instead of \(T_{\text{double}}\)

We thus have:

\(
\text{new value} = \text{initial value} \times \left(\frac{1}{2}\right)^{t/T_{\text{half}}}
\)

parallel to

\(
\text{new value} = \text{initial value} \times 2^{t/T_{\text{double}}}
\)

Exercise 44, p490

The half-life of a drug in the bloodstream is 4 hours. What fraction of the original drug dose remains in 24 hours? In 48 hours?

Solution 1.
Half of the drug remains after 4 hours;
half of this half, which is a quarter after 8 hours,
half of his quarter, which is \(\frac{1}{8}\) after 12 hours;
another 12 hours will leave an eighth of this \(\frac{1}{8}\);
We thus have \( \frac{1}{64} \) after 24 hours;

Another 24 hours will leave the bloodstream with \( \frac{1}{64} \) of this \( \frac{1}{64} \);
Thus after 48 hours the there is \( \frac{1}{64} \times \frac{1}{64} =\frac{1}{4096} \) of the initial quantity;

The half-life of a drug in the bloodstream is 4 hours. What fraction of the original drug dose remains in 24 hours? In 48 hours?

Solution 2.
We us the formula with \(T_{\text{half}}=4\):

\(
\text{new value} = \text{initial value} \times \left(\frac{1}{2}\right)^{t/T_{\text{half}}}
\)

For t=24 hours, the factor is

\( \left(\frac{1}{2}\right)^{t/T_{\text{half}}} = \left(\frac{1}{2}\right)^{24/4} = \left(\frac{1}{2}\right)^6 = \frac{1}{64} \).

For t=48 hours, the factor is

\( \left(\frac{1}{2}\right)^{t/T_{\text{half}}} = \left(\frac{1}{2}\right)^{48/4} = \left(\frac{1}{2}\right)^{12} = \frac{1}{4096} \).

Is exactly the same as for doubling time:

If \(P\) is the percentage decay rate, then

\(
T_{\text{half}} \approx \frac{70}{P}
\)

assuming that \(P\) is relatively small.

Note that \(P\) here is in __ percents __ per unit of time.

One makes use of this formula in a way similar to that of doubling time formula.

\(T_{\text{double/half}} = \frac{\log_{10}2}{\log_{10}(1+r)} \)

where \(r\) is the rate of change (in decimal!)

This is the rate of growth if \(r>0\), and

the rate of decay if \(r<0\).

The formula calculates \(T_{\text{double}} \) for positive \(r\) (growth), and

it calculates \(T_{\text{half}} \) for negative \(r\) (decay)

\(a=\log_{10}x\) means nothing but \(10^a=x\)

Example. \(2=\log_{10}100 \) because \(10^2=100\)

Useful rules to apply

\(\log_{10}10^x=x\)

\(10^{\log_{10}x} =x \ \ \ (x>0)\)

\(\log_{10}(xy)=\log_{10}x+\log_{10}y \ \ \ (x>0 \ \text{and} \ y>0)\)

\(\log_{10}a^x=x\log_{10}a \ \ \ (a>0) \)

More Details in Brief Review on p488 of the textbook.