## Review for the third midterm

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

#### Two basic growth patterns

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.

#### Doubling time and half-times formulas

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!)

##### Exercise 15 p.479 (modified)

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 ##### Exercise 15 p.479 (modified) ... continued 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 $$100,000\times 1.02^{20} \approx \148,595$$ ##### Exercise 15 p.479 (modified) ... variation 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. ##### Exercise 15 p.479 (modified) ... continued 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 ). ##### Exercise 15 p.479 (modified) ... continued ... summary 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. ##### Exercise 21 p. 480 magic penny 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.

##### Exercise 21 p. 480 magic penny ... continued

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.

#### Digression on Logarithms

$$a=\log_{10}x$$ 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.

#### Logarithms Exercises 13,16,17,19

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 $$\log_{10} 1,600,000= \log_{10} (1.6 \times 10^6)=6+\log_{10}1.6$$.

#### Logarithms Exercise 23

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

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

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

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

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

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

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

#### Doubling time. Exercise 29

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 $$12000 \times 2^{24/10} = 63,336$$

###### Doubling Time. Exercise 33

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 $$\approx 20.6 \ \text{billion}$$ $$\text{new value} = \text{initial value} \times 2^{t/T_{\text{double}}} = 7.1 \times 2^{100/45} \approx 33.1 \ \text{billion}$$

#### Doubling Time. Exercise 36 (modified)

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

#### Doubling Time. Exercise 40

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

#### Doubling Time. Exercise 40 ... continued

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 $$1.24$$

#### Half-Life. Exercise 46

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

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

#### Half-Life. Exercise 52

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

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

#### Half-life Exercise 52 ... continued

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 $$\text{new value} = \text{initial value} \times \left(\frac{1}{2}\right)^{t/T_{\text{half}}} = 5000 \times \left(\frac{1}{2}\right)^{10/14} \approx3050$$

#### Exact half-life/doubling time. Exercise 56

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 $$\text{new value} = \text{initial value} \times 2^{t/T_{\text{double}}} = 100 \times 2^{52/3.8} \approx 1.3 \ \text{million}$$