# Section 8B

### exercises

...

#### Logarithms Exercises 14,16,18,20

Determine true or false without calculations

$$10^{3.334}$$ between $$500$$ and $$1000$$ false because $$10^{3.334} > 10^3=1000$$

$$10^{-5.2}$$ between $$-100,000$$ and $$-1,000,000$$ false because $$10^{-5.2} > 0$$

$$\log_{10}96$$ between $$3$$ and $$4$$ false because $$10^3=1000$$

$$\log_{10}(8 \times 10^9)$$ between $$9$$ and $$10$$ true because $$\log_{10}(8 \times 10^9)= 9+\log_{10}8$$,
and $$\log_{10}8$$ is between $$0$$ and $$1$$.

#### Logarithms Exercise 24

Using approximation $$\log_{10}5 \approx 0.699$$, evaluate each of the following without calculator.

$$\log_{10}50 = \log_{10}5 + \log_{10}10 \approx 0.699 +1 = 1.669$$

$$\log_{10}5000 = \log_{10}5 + \log_{10}1000 \approx 0.699 +3 = 3.669$$

$$\log_{10}0.05 = \log_{10}5 + \log_{10}10^{-2} \approx 0.699 -2 = -1.301$$

$$\log_{10}25 = \log_{10}5^2 =2\log_{10}5 \approx 2 \times 0.699 = 1.398$$

$$\log_{10}0.20 = \log_{10}5^{-1} =-\log_{10}5 \approx -0.699$$

$$\log_{10}0.04 = \log_{10}5^{-2} =-2 \times \log_{10}5 \approx -1.398$$

#### Doubling time Exercise 28

Prices are rising with a doubling time of 2 months. By what factor do prices increase in a year?

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

With $$T_{\text{double}} = 2$$, and $$t=12$$ the factor in question is

$$2^{t/T_{\text{double}} }= 2^{12/2} =2^6 = 64$$.

#### Doubling time Exercise 30

The initial population in a town is $$12,000$$, and it grows with a doubling time of $$8$$ 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 $$T_{\text{double}} = 8$$, initial value of $$12,000$$, and $$t=12$$ we calculate

$$\text{new value} = 12,000 \times 2^{12/8} \approx 33,941$$

While for $$t=24$$ we calculate

$$\text{new value} = 12,000 \times 2^{24/8} \approx 96,000$$

#### Doubling time Exercise 34

In mid-2013, estimated world population was 7.1 billion. Assume the doubling time of 65 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 $$T_{\text{double}} = 65$$ years.

For 2023, $$t=10$$ years, and the prediction is $$\text{new value} = \text{initial value} \times 2^{t/T_{\text{double}}} = 7.1 \times 2^{10/65} \approx 7.9 \ \text{billion}$$

For 2063, $$t=50$$ years, and the prediction is $$\text{new value} = \text{initial value} \times 2^{t/T_{\text{double}}} = 7.1 \times 2^{50/65} \approx 12.1 \ \text{billion}$$

For 2113, $$t=100$$ years, and the prediction is $$\approx 20.6 \ \text{billion}$$ $$\text{new value} = \text{initial value} \times 2^{t/T_{\text{double}}} = 7.1 \times 2^{100/65} \approx 20.6 \ \text{billion}$$

#### Doubling time Exercise 38

A city's population is growing at a rate of 3.5% per year. What is its doubling time? By what factor will the population increase in 50 years?

Solution. The approximate doubling time formula gives us the doubling time of

$$T_{\text{double}} \approx \frac{70}{P} = \frac{70}{3.5} =20 \ \text{years}$$

We us the formula $$\text{new value} = \text{initial value} \times 2^{t/T_{\text{double}}}$$

with $$T_{\text{double}} = 20$$ years and $$t=50$$ years to calculate the factor as $$2^{t/T_{\text{double}}} = 2^{50/20} \approx 5.7$$

#### Half-life Exercise 42

The half-life of a radioactive substance is 400 years. If you start with some amount of this substance, what fraction will remain in 120 years? in 2500 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 $$T_{\text{half}} = 400$$ years.

The remaining fraction after $$t=120$$ years is the factor

$$\left(\frac{1}{2}\right)^{t/T_{\text{half}}} = \left(\frac{1}{2}\right)^{120/400} \approx 0.81 = 81\%$$

Similarly, after $$t=2500$$ years, $$\left(\frac{1}{2}\right)^{t/T_{\text{half}}} = \left(\frac{1}{2}\right)^{2500/400} \approx 0.013 = 1.3\%$$

#### Half-life Exercise 48

Radium-226 is a metal with a half-life of 1600 years. If you start with 1 kilogram of radium-226, how much will remain after 1000 years? after 10,000 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 $$T_{\text{half}} = 1600$$ years and the initial value of 1 kilogram.

$$t=1000$$ years later we have:

$$\text{new value} = 1 \times \left(\frac{1}{2}\right)^{1000/1600} \approx 0.648 = 648 \ \text{gram}$$

$$t=10,000$$ years later we have:

$$\text{new value} = 1 \times \left(\frac{1}{2}\right)^{10000/1600} \approx 0.013 = 13 \ \text{gram}$$

#### Half-life Exercise 50

A clean-up project is reducing the concentration of a pollutant in the water supply with a 3.5% decrease per week. What is the approximate half-life of the concentration of the pollutant? About what fraction of the original amount of the pollutant will remain when the project ends after one year (52 weeks)?

Solution. We determine the approximate half-life as

$$T_{\text{half}} \approx \frac{70}{P} = \frac{70}{3.5} =20 \ \text{week}$$

#### Half-life Exercise 50 ... continued

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

with $$t=52$$ and $$T_{\text{half}} = 20$$ to find the factor

$$\left(\frac{1}{2}\right)^{t/T_{\text{half}}} = \left(\frac{1}{2}\right)^{52/20} \approx 0.165 = 16.5\%$$

#### Exercise 54

Hyperinflation is driving up prices at a rate of 80% per month. For an item that costs $$\1,000$$ today, what will be the price in 1 year?

Solution 1 We firstly use the exact doubling time formula:

$$T_{\text{double}} = \frac{\log_{10}2}{\log_{10}(1+r)} = \frac{\log_{10}2}{\log_{10}(1.8)} \approx 1.17924958484 \ \text{month}$$

We now use this value, $$t=12$$, and initial value of $$\1000$$ in the usual formula:

$$\text{new value} = \text{initial value} \times 2^{t/T_{\text{double}}} \\ = 1000 \times 2^{12/1.17924958484} \approx \ 1,156,831$$

#### Exercise 54 ... continued

Hyperinflation is driving up prices at a rate of 80% per month. For an item that costs $$\1,000$$ today, what will be the price in 1 year?

Solution 2. Observe that every month the price of this item is multiplied by a factor of $$1.8$$. After $$12$$ months, the price is thus multiplied by this factor $$12$$ times. Equivalently, it is multiplied by a factor of $$1.8^{12}$$. Thus the price after a year will be $$1000 \times 1.8^{12} \approx \ 1,156,831$$