Operations on the set of real numbers

Unit 1 Operations on the set of real numbers

Key unit competence: Apply operations on real numbers, radicals and logarithms in solving mathematical and real-life problems.

Key ideas

• Absolute value: $|x|=x$ if $x\ge 0$ and $|x|=-x$ if $x<0$. It is the distance from $x$ to 0. Properties: $|xy|=|x||y|$, $\left|\frac{x}{y}\right|=\frac{|x|}{|y|}$, $|x+y|\le |x|+|y|$ (triangle inequality), $|x|=a⟺x=\pm a$ ($a\ge 0$).
• Radicals: $\sqrt[n]{a^m}=a^{\frac{m}{n}}$; $\sqrt[n]{ab}=\sqrt[n]{a}\sqrt[n]{b}$. Rationalise denominators with the conjugate: $(\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b})=a-b$.
• Decimal logarithm: $\log x=y⟺10^y=x$ ($x>0$). $\log 1=0$, $\log 10=1$, $\log 10^n=n$.
• Properties: $\log (xy)=\log x+\log y$, $\log \frac{x}{y}=\log x-\log y$, $\log x^n=n\log x$.
• Models: linear growth $y=a+bt$; exponential growth $y=a{b}^t$ ($b>1$); exponential decay $y=a{b}^t$ ($0<b<1$); compound interest $A=P{\left(1+\frac{r}{n}\right)}^{nt}$. Logarithms solve for the time: $t=\frac{\log (y/a)}{\log b}$.
• Earthquakes (Richter scale): $M=\log \frac{I}{I_0}$. One unit more in magnitude means 10 times the intensity.

Formulas to know

Common misconceptions

• "$|x-3|=x-3$ for all $x$." No: only when $x\ge 3$; otherwise it is $3-x$.
• "$\log (x+y)=\log x+\log y$." No: the product rule is $\log (xy)=\log x+\log y$.
• "A magnitude 7 earthquake is twice as strong as magnitude 5." No: it is $10^2=100$ times as intense.

Worked examples

Practice questions

Easier and harder questions are mixed, just like in a real exam.

1. Solve $|3x+2|=7$.

   Show answer

   $3x+2=7$ gives $x=\frac{5}{3}$; $3x+2=-7$ gives $x=-3$.

   Common mistake: Forgetting the negative case.

2. A town has 50 000 people, growing by 3% per year. Write a model and find the population after 10 years.

   Show answer

   $P=50000\times 1.03^t$; $P(10)\approx 67196$.

   Common mistake: Using the linear model $50000+0.03t$.

3. Simplify $\log 4+2\log 5-\log 10$.

   Show answer

   $\log 4+\log 25-\log 10=\log \frac{100}{10}=\log 10=1$.

   Common mistake: Writing $2\log 5=\log 10$.

4. Prove that $|x|\le a⟺-a\le x\le a$ for $a\ge 0$, and use it to solve $|x-1|<3$.

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   Both say the distance from $x$ to 0 is at most $a$. $-3<x-1<3$, so $-2<x<4$.

   Common mistake: Writing $x-1<\pm 3$.

5. Rationalise $\frac{\sqrt{2}}{\sqrt{6}-\sqrt{2}}$.

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   $\frac{\sqrt{2}(\sqrt{6}+\sqrt{2})}{6-2}=\frac{\sqrt{12}+2}{4}=\frac{2\sqrt{3}+2}{4}=\frac{\sqrt{3}+1}{2}$.

   Common mistake: Multiplying by $\sqrt{6}-\sqrt{2}$ again.

6. Solve $\log (2x+1)=2$.

   Show answer

   $2x+1=100$, $x=49.5$.

   Common mistake: Writing $2x+1=20$.

7. The value of a car is $V=12000000\times 0.85^t$ FRW. Find its value after 3 years and when it first falls below 6 000 000 FRW.

   Show answer

   $V(3)=12000000\times 0.614125\approx 7369500$ FRW. $0.85^t<0.5\Rightarrow t>\frac{\log 0.5}{\log 0.85}\approx 4.27$, so it is below 6 000 000 FRW after about 4.3 years (from the 5th year on).

   Common mistake: Dividing logs of the wrong numbers.

8. The 2008 Sichuan earthquake had magnitude 7.9 and a smaller one 5.9. How many times more intense was the first?

   Show answer

   $10^{7.9-5.9}=10^2=100$ times.

   Common mistake: Taking the ratio $\frac{7.9}{5.9}$.

Matching and drag-and-drop

1. Matching

   Match each item in column A with the correct answer in column B. Write the letter next to each number.

   Column A	Column B
   1. Given $\log 2\approx 0.301$ and $\log 3\approx 0.477$, find $\log 18$.	a. $x\le -\frac{4}{3}$ or $x\ge 2$
   2. Earthquakes of magnitude 7.1 and 5.6 are compared. The first is how many times more intense?	b. $1.255$
   3. Rationalise $\frac{3}{2-\sqrt{3}}$.	c. $6+3\sqrt{3}$
   4. Solve $|3x-1|\ge 5$.	d. $\approx 31.6$

   Show answer

   1 → b ($1.255$) · 2 → d ($\approx 31.6$) · 3 → c ($6+3\sqrt{3}$) · 4 → a ($x\le -\frac{4}{3}$ or $x\ge 2$)

   • Step 1. $18=2\times 3^2$.
     Step 2. $\log 18=\log 2+2\log 3$.
     Step 3. $0.301+2(0.477)=1.255$.
   • Step 1. $M_1-M_2=\log \frac{I_1}{I_2}=1.5$.
     Step 2. $\frac{I_1}{I_2}=10^{1.5}\approx 31.6$.
   • Step 1. Multiply by $\frac{2+\sqrt{3}}{2+\sqrt{3}}$.
     Step 2. Denominator: $(2-\sqrt{3})(2+\sqrt{3})=4-3=1$.
     Step 3. Result: $3(2+\sqrt{3})=6+3\sqrt{3}$.
   • Step 1. $3x-1\ge 5$ gives $x\ge 2$.
     Step 2. $3x-1\le -5$ gives $3x\le -4$, so $x\le -\frac{4}{3}$.
     Step 3. "Greater than" gives two separate parts.

2. Drag and drop

   Choose the correct word or phrase from the box for each gap. One word in the box is not needed.

   $7.5$  ·  $112683$ FRW  ·  $101000$ FRW  ·  $4913000$ FRW
   1. 100 000 FRW is invested at 12% per year compounded monthly. The amount after 1 year is about .
   2. A car worth 8 000 000 FRW loses 15% of its value each year. Its value after 3 years is .
   3. Earthquake A is 1 000 times as intense as earthquake B, which has magnitude 4.5. The magnitude of A is .
   Show answer

   1: $112683$ FRW · 2: $4913000$ FRW · 3: $7.5$

   • Step 1. Monthly rate $\frac{12\%}{12}=1\%$, 12 periods.
     Step 2. $A=100000\times 1.01^{12}\approx 100000\times 1.12683=112683$ FRW.
   • Step 1. Each year multiply by $1-0.15=0.85$.
     Step 2. $8000000\times 0.85^3=8000000\times 0.614125=4913000$ FRW.
   • Step 1. $M_A-M_B=\log \frac{I_A}{I_B}=\log 1000=3$.
     Step 2. $M_A=4.5+3=7.5$.

3. Matching

   Match each item in column A with the correct answer in column B. Write the letter next to each number.

   Column A	Column B
   1. Simplify $|\sqrt{2}-2|+|\sqrt{2}-1|$.	a. $10800$
   2. Solve $10^{2x-1}=1000$.	b. $-7<x<1$
   3. Solve $|x+3|<4$.	c. $x=2$
   4. A culture of 400 bacteria triples every 2 hours. How many are there after 6 hours?	d. $1$

   Show answer

   1 → d ($1$) · 2 → c ($x=2$) · 3 → b ($-7<x<1$) · 4 → a ($10800$)

   • Step 1. $\sqrt{2}\approx 1.41<2$, so $|\sqrt{2}-2|=2-\sqrt{2}$.
     Step 2. $\sqrt{2}>1$, so $|\sqrt{2}-1|=\sqrt{2}-1$.
     Step 3. $(2-\sqrt{2})+(\sqrt{2}-1)=1$.
   • Step 1. $1000=10^3$.
     Step 2. $2x-1=3$, so $x=2$.
   • Step 1. $|x+3|<4⟺-4<x+3<4$.
     Step 2. Subtract 3 from every part: $-7<x<1$.
   • Step 1. 6 hours $=3$ periods of 2 hours.
     Step 2. $N=400\times 3^3=400\times 27=10800$.

4. Drag and drop

   Choose the correct word or phrase from the box for each gap. One word in the box is not needed.

   10 years  ·  $0.051$  ·  $2$  ·  about 7.3 years
   1. Given $\log 2\approx 0.301$ and $\log 3\approx 0.477$, find $\log \frac{9}{8}$. →
   2. How long does 500 000 FRW take to double at 10% per year compound interest? →
   3. Without a calculator, find $\log 25+\log 40-\log 10$. →
   Show answer

   1: $0.051$ · 2: about 7.3 years · 3: $2$

   • Step 1. $\log \frac{9}{8}=2\log 3-3\log 2$.
     Step 2. $0.954-0.903=0.051$.
   • Step 1. $500000\times 1.1^t=1000000$, so $1.1^t=2$.
     Step 2. $t=\frac{\log 2}{\log 1.1}=\frac{0.3010}{0.0414}\approx 7.27$ years.
   • Step 1. $\log 25+\log 40=\log (25\times 40)=\log 1000$.
     Step 2. $\log 1000-\log 10=\log 100$.
     Step 3. $\log 100=2$.

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