... Sections 8A,8B ...

Linear growth occurs when a quantity grows by the same __absolute__ amount in each unit of time

Exponential growth occurs when a quantity grows by the same __relative__ amount in each unit of time

Same definitions of linear and exponential apply to decay instead of growth.

Approximate:

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

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

assuming that \(P\) is relatively small.

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

Exact:

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

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

The value of your house is increasing by \( \$2,000 \) per year. If it is worth \( \$100,000 \) today, what will it be worth in twenty years?

Solution.

This is a __linear__ growth.

Every year the value increases by $2,000. Thus over the period of ten years the value increases by

\(\$ 2,000 \times 20 = \$40,000\)

The value becomes

\(100,000 + 40,000 = \$140,000.\) <===== answer

The value of your house is increasing by 2% per year. If it is worth $100,000 today, what will it be worth in twenty years?

Solution.

This is a __exponential__ growth.

Every year the value increases by 2% of what it was.

The value after 1 year is \(100,000+2,000 \times 0.02=100,000 \times (1+0.02)=100000 \times 1.02. \)

The value after 2 years is \(100,000\times 1.02 + 100,000 \times 1.002 \times 0.02 =100000 \times 1.02^2. \)

The value after 3 years is \(100,000\times 1.02^3\)

The value of your house is increasing by 2% per year. If it is worth $100,000 today, what will it be worth in twenty years?

Solution.

This is a __exponential__ growth, and the pattern is clear now.

The value after 20 years is \(100,000\times 1.02^{20} \approx \$148,595\)

The value of your house is increasing by \( \$2,000 \) per year. It is worth \( \$100,000 \) today, how long will it take for its value to double?

Solution.

This is a __linear__ growth, and __ no __ doubling time formula applies.

Every year the value increases by \( $2,000 \). We want the value to increase by another \(\$100,000 \).

With a \( \$2,000 \) per year growth it will take

\(\frac{100,000}{2,000} = 50\) years for the value to double.

The value of your house is increasing by 2% per year. It is worth \( \$100,000 \) today, how long will it take for its value to double?

Solution.

This is a __exponential__ growth, and doubling time formula applies.

We can calculate approximately:

\(
T_{\text{double}} \approx \frac{70}{P} = \frac{70}{2} = 35
\) years.

We can calculate exactly:

\(T_{\text{double/half}}= \frac{\log_{10}2}{\log_{10}(1+r)}= \frac{\log_{10}2}{\log_{10}(1+0.02)} \approx 35 \) years ( 35.003 ).

The house which is worth \( \$100,000 \) today increases in value either by $2\%$ or by \($2,000\) per year.

One year later the value of the house is \( \$102,000 \) in either case.

Twenty years later the value of the house is \( \$140,000 \) versus \(\$148,595\). Exponential growth wins slightly.

The value doubles (i.e. reaches \( \$200,000 \)) in \(50\) years with linear growth, while only in \(35\) years with exponential growth.

Recall: $1$ penny on day zero, $2$ pennies on day $1$, and so further...

How many days would elapse before you have a total of more than \(\$ 1\) billion?

Solution. The pattern: \( 2^n \) pennies after $n$ days elapsed.

Recall: \(\$ 1\) billion = \(\$ 10^9 = 10^9 \times 100 = 10^{11} \) pennies

You may calculate by trial and error as the book suggests:

\(2^{37}=137438953472>10^{11}\) while \(2^{36}=68719476736 <10^{11} \)

Answer: \(37\) days.

How many days would elapse before you have a total of more than \(\$ 1\) billion?

Alternative solution.

We want the smallest integer \(n\) such that \(2^n > 10^{11} \)

Take logarithms: \( \log_{10} 2^n > \log_{10} \left( 10^{11} \right) \)

Transforms to \(n \times \log_{10} 2 > 11\)

That is \(n > \frac{11}{\log_{10} 2} \approx 36.5 \)

Answer: \(37\) days.

\(a=\log_{10}x\) 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.

Determine true or false without calculations

\(10^{0.928}\) is between \(1\) and \(10\) true because \(10^0=1 \) and \(10^1=10\)

\(10^{-2.67}\) is between \(0.001\) and \(0.01\) true because \(10^{-2} = 0.01 \) and \(10^{-3}=0.001\)

\(\log_{10}\pi\) is between \(3\) and \(4\) false because \(10^3=1000\) and \(10^4=10000\) while \(\pi = 3.1415\ldots\)

\(\log_{10}1,600,000 \) is between \(16\) and \(17\) false because \(\log_{10} 1,600,000= \log_{10} (1.6 \times 10^6)=6+\log_{10}1.6\).

Using approximation \(\log_{10}2 \approx 0.301 \), evaluate each of the following without calculator.

\(\log_{10}8 = \log_{10}2^3=3\log_{10} 2 \approx 3 \times 0.301= 0.903\)

\(\log_{10}2000 = \log_{10}2 + \log_{10}1000 \approx 0.301 +3 = 3.301 \)

\(\log_{10}0.5 = \log_{10} \frac{1}{2} = \log_{10}2^{-1} = - \log_{10}2 \approx -0.301\)

\(\log_{10}64 = \log_{10}2^6 =6\log_{10}2 \approx 6 \times 0.301 = 1.806 \)

\(\log_{10}1/8 = \log_{10}2^{-3} =-3\log_{10}2 \approx -3 \times 0.301 = -0.903 \)

\(\log_{10}0.2 = \log_{10}\left(2 \times 10^{-1}\right) = \log_{10}2 -1 \approx 0.301-1 = -0.699 \)

The initial population of a town is \(12,000\), and it grows with a doubling time of \(10\) years. What will the population be in \(12\) years? in \(24\) years?

Solution.

We use the formula

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

with \(T_{\text{double}} = 10\), and the initial value of \(12,000\)

Thus in \(12\) years we have \( 12000 \times 2^{12/10} = 27,568 \)

after \(24\) years we have \( 12000 \times 2^{24/10} = 63,336 \)

In mid-2013, estimated world population was 7.1 billion. Assume the doubling time of 45 years to predict the population in 2023,2063, and 2113.

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

with the initial value of 7.1 billion and \(T_{\text{double}} = 45\) years.

For 2023, \(t=10\) years, and the prediction is \( \text{new value} = \text{initial value} \times 2^{t/T_{\text{double}}} = 7.1 \times 2^{10/45} \approx 8.3 \ \text{billion} \)

For 2063, \(t=50\) years, and the prediction is \( \text{new value} = \text{initial value} \times 2^{t/T_{\text{double}}} = 7.1 \times 2^{50/45} \approx 15.3 \ \text{billion} \)

For 2113, \(t=100\) years, and the prediction is \(\approx 20.6 \ \text{billion}\) \( \text{new value} = \text{initial value} \times 2^{t/T_{\text{double}}} = 7.1 \times 2^{100/45} \approx 33.1 \ \text{billion} \)

A community of mice begins with an initial population of \(1000\) and grows \(10\%\) per month. What is the approximate doubling time?

Solution. We use the formula \( T_{\text{double}} \approx \frac{70}{P} \)

to find

\(
T_{\text{double}} \approx \frac{70}{10} = 7 \ \text{months}
\)

Oil consumption is increasing at a rate of \(2.2\%\) per year. What is its doubling time?

Solution. We use the formula \( T_{\text{double}} \approx \frac{70}{P} \)

to find

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

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

Solution. We have already calculated that \( T_{\text{double}} \approx 32 \) years, and now use the formula

\(
\text{new value} = \text{initial value} \times 2^{t/T_{\text{double}}} = \text{initial value} \times 2^{10/32} \)

\(= \text{initial value} \times 1.24
\)

Answer: by a factor of \(1.24\)

The current population of a threatened animal species is \(1\) million, but it is declining with a half-life of \(25\) years. How many animals will be left in \(70\) years?

Solution. We use the half-life formula \( \text{new value} = \text{initial value} \times \left(\frac{1}{2}\right)^{t/T_{\text{half}}} \)

with \(T_{\text{half}} = 25\) years, and calculate

\( \text{new value} = 1 \times \left(\frac{1}{2}\right)^{70/25} \approx 0.144 \ \text{million} = 144 \ \text{thousand animals} \)

The production of a gold mine decreases by \(5\%\) per year. What is the approximate half-life for the production decline?

Solution. We determine the approximate half-life as

\( T_{\text{half}} \approx \frac{70}{P} = \frac{70}{5} =14 \ \text{years} \)

The production of a gold mine decreases by \(5\%\) per year. If its current production is \(5,000\) kilograms, about what will its production be in \(10\) years?

We have already calculated that \( T_{\text{half}} \approx 14 \ \text{years}\) and now use the formula \( \text{new value} = \text{initial value} \times \left(\frac{1}{2}\right)^{t/T_{\text{half}}} = 5000 \times \left(\frac{1}{2}\right)^{10/14} \approx3050 \)

A family of \(100\) termites invades your home, and its population increases at a rate of \(20\%\) per week. How many termites will be in your home after \(1\) year (\(52\) weeks)?

Solution. We firstly use the exact doubling time formula:

\( T_{\text{double}} = \frac{\log_{10}2}{\log_{10}(1+r)} = \frac{\log_{10}2}{\log_{10}(1.2)} \approx 3.8 \ \text{weeks} \)

We now use the formula \( \text{new value} = \text{initial value} \times 2^{t/T_{\text{double}}} = 100 \times 2^{52/3.8} \approx 1.3 \ \text{million} \)