Answer Key

Exponents and Indices

Pack A — Answers

#	Question	Answer
1	Evaluate $2^{5}$.	32
2	Evaluate $7^{0}$.	1
3	Simplify $x^{3} \times x^{4}$.	$x^7$
4	Simplify $\dfrac{x^{8}}{x^{3}}$.	$x^5$
5	Simplify $(x^{3})^{2}$.	$x^6$
6	Simplify $(2 x^{3})^2$.	$4 x^6$
7	Write $2^{-3}$ as a fraction.	$\dfrac{1}{8}$
8	Write $3000$ in standard form (scientific notation).	$3 \times 10^3$
9	Write $0.004$ in standard form.	$4 \times 10^{-3}$
10	Write $2.5 \times 10^{4}$ as an ordinary number.	25 000
11	Simplify $3 x^{2} \times 4 x^{5}$.	$12 x^7$
12	Simplify $\dfrac{20 x^{7}}{4 x^{3}}$.	$5 x^4$
13	Simplify $\left(\dfrac{x^{5}}{x^{2}}\right)^{3}$.	$x^9$
14	Simplify $x^{5} \times x^{-2}$.	$x^3$
15	Write $\dfrac{1}{x^{4}}$ using a negative index.	$x^{-4}$
16	Calculate $(3 \times 10^{4}) \times (2 \times 10^{5})$, giving your answer in standard form.	$6 \times 10^9$
17	Calculate $\dfrac{8 \times 10^{7}}{2 \times 10^{3}}$ in standard form.	$4 \times 10^4$
18	Simplify $\dfrac{x^{4} \times x^{3}}{x^{2}}$.	$x^5$
19	How many times larger is $6 \times 10^{7}$ than $2 \times 10^{3}$?	$3 \times 10^4$ (30 000 times)
20	Simplify $(-3 x^{2})^{2}$.	$9 x^4$
21	Evaluate $25^{1/2}$.	5
22	Evaluate $27^{1/3}$.	3
23	Evaluate $8^{2/3}$.	4
24	Evaluate $16^{-1/2}$.	$\dfrac{1}{4}$
25	Simplify $\dfrac{(2 x^{3})^{2}}{x^{4}}$.	$4 x^2$
26	Calculate $(3 \times 10^{5}) + (5 \times 10^{4})$ in standard form.	$3.5 \times 10^5$
27	Simplify $(8)^{-2/3}$.	$\dfrac{1}{4}$
28	Solve $2^x = 32$.	$x = 5$
29	If $x = \dfrac{1}{2}$, evaluate $x^{4}$.	$\dfrac{1}{16}$
30	Write $\sqrt[3]{x^{2}}$ using a fractional index.	$x^{2/3}$
31	Solve $2^{x+1} = 32$.	$x = 4$
32	Solve $3^{2x} = 81$.	$x = 2$
33	Simplify $\dfrac{(2 x^{3})^{2} \cdot x^{5}}{x^{4}}$.	$4 x^7$
34	The mass of Earth is $6 \times 10^{24}$ kg and the mass of the Sun is $2 \times 10^{30}$ kg. How many times more massive is the Sun?	$\approx 3.33 \times 10^5$ times (≈ 333 000)
35	Simplify $\dfrac{x^{3} \cdot y^{-2}}{x^{-1} \cdot y^{4}}$, expressing the answer with positive indices.	$\dfrac{x^4}{y^6}$
36	Evaluate $(16)^{3/4}$.	8
37	Light travels at $3 \times 10^8$ m/s. Calculate the distance light travels in $600$ seconds, in standard form.	$1.8 \times 10^{11}$ m
38	Solve $2^{x} \cdot 2^{x+1} = 32$.	$x = 2$
39	Simplify $\left(\dfrac{2x^{5}y^{3}}{x^{2}y^{1}}\right)^{2}$.	$4 x^6 y^4$
40	Solve $x^{1/3} = 4$ for positive $x$.	$x = 64$

Pack B — Answers

#	Question	Answer
1	Evaluate $3^{4}$.	81
2	Evaluate $12^{0}$.	1
3	Simplify $x^{5} \times x^{2}$.	$x^7$
4	Simplify $\dfrac{x^{10}}{x^{4}}$.	$x^6$
5	Simplify $(x^{4})^{3}$.	$x^{12}$
6	Simplify $(5 x^{2})^2$.	$25 x^4$
7	Write $5^{-2}$ as a fraction.	$\dfrac{1}{25}$
8	Write $45000$ in standard form (scientific notation).	$4.5 \times 10^4$
9	Write $0.00007$ in standard form.	$7 \times 10^{-5}$
10	Write $6.3 \times 10^{3}$ as an ordinary number.	6 300
11	Simplify $2 x^{3} \times 5 x^{4}$.	$10 x^7$
12	Simplify $\dfrac{18 x^{8}}{3 x^{2}}$.	$6 x^6$
13	Simplify $\left(\dfrac{x^{7}}{x^{3}}\right)^{2}$.	$x^8$
14	Simplify $x^{7} \times x^{-4}$.	$x^3$
15	Write $\dfrac{1}{x^{6}}$ using a negative index.	$x^{-6}$
16	Calculate $(4 \times 10^{3}) \times (5 \times 10^{2})$, giving your answer in standard form.	$2 \times 10^6$
17	Calculate $\dfrac{9 \times 10^{5}}{3 \times 10^{2}}$ in standard form.	$3 \times 10^3$
18	Simplify $\dfrac{x^{6} \times x^{2}}{x^{5}}$.	$x^3$
19	How many times larger is $6 \times 10^{8}$ than $2 \times 10^{4}$?	$3 \times 10^4$ (30 000 times)
20	Simplify $(-2 x^{3})^{3}$.	$-8 x^9$
21	Evaluate $81^{1/2}$.	9
22	Evaluate $64^{1/3}$.	4
23	Evaluate $27^{2/3}$.	9
24	Evaluate $49^{-1/2}$.	$\dfrac{1}{7}$
25	Simplify $\dfrac{(3 x^{2})^{3}}{x^{4}}$.	$27 x^2$
26	Calculate $(4 \times 10^{6}) + (2 \times 10^{5})$ in standard form.	$4.2 \times 10^6$
27	Simplify $(27)^{-2/3}$.	$\dfrac{1}{9}$
28	Solve $2^x = 128$.	$x = 7$
29	If $x = \dfrac{2}{3}$, evaluate $x^{3}$.	$\dfrac{8}{27}$
30	Write $\sqrt[5]{x^{3}}$ using a fractional index.	$x^{3/5}$
31	Solve $2^{x+1} = 64$.	$x = 5$
32	Solve $3^{2x} = 729$.	$x = 3$
33	Simplify $\dfrac{(3 x^{2})^{2} \cdot x^{3}}{x^{5}}$.	$9 x^2$
34	The mass of Earth is $6 \times 10^{24}$ kg and the mass of the Sun is $2 \times 10^{30}$ kg. How many times more massive is the Sun?	$\approx 3.33 \times 10^5$ times (≈ 333 000)
35	Simplify $\dfrac{x^{4} \cdot y^{-3}}{x^{-2} \cdot y^{5}}$, expressing the answer with positive indices.	$\dfrac{x^6}{y^8}$
36	Evaluate $(81)^{3/4}$.	27
37	Light travels at $3 \times 10^8$ m/s. Calculate the distance light travels in $200$ seconds, in standard form.	$6 \times 10^{10}$ m
38	Solve $2^{x} \cdot 2^{x+1} = 128$.	$x = 3$
39	Simplify $\left(\dfrac{3x^{4}y^{5}}{x^{1}y^{2}}\right)^{2}$.	$9 x^6 y^6$
40	Solve $x^{1/3} = 5$ for positive $x$.	$x = 125$

Problems — Worked Solutions

**Powers of 2.** A grain of rice is placed on the first square of a chessboard. The number of grains doubles on each successive square. (a) How many grains are on square 1, 2, 3, 4, 5? (b) Write a formula for the number of grains on square $n$. (c) How many grains are on square 32, in standard form (3 s.f.)?

Answer

(a) 1, 2, 4, 8, 16 (b) $g_n = 2^{n-1}$ (c) $\approx 2.15 \times 10^9$

(a) Doubling sequence: 1, 2, 4, 8, 16. (b) Geometric with $u_1 = 1$, $r = 2$. So $g_n = 1 \cdot 2^{n-1} = 2^{n-1}$. (c) $g_{32} = 2^{31} = 2{,}147{,}483{,}648 \approx \mathbf{2.15 \times 10^9}$.

**Cell biology.** A red blood cell has a diameter of about $7.5 \times 10^{-6}$ m. A human capillary has a diameter of about $9 \times 10^{-6}$ m. (a) How many times wider is the capillary than the red blood cell? Give your answer to 2 s.f. (b) Can the red blood cell fit through the capillary? Justify. (c) If a person has approximately $5 \times 10^{12}$ red blood cells, calculate the total volume in m³, treating each cell as a sphere with the diameter above. Use $V = \frac{4}{3}\pi r^3$ and give the answer in standard form (2 s.f.).

Answer

(a) 1.2 times (b) Yes, just (c) $\approx 1.1 \times 10^{-3}$ m³ (= 1.1 L)

(a) $\frac{9 \times 10^{-6}}{7.5 \times 10^{-6}} = \frac{9}{7.5} = 1.2$. (b) The capillary (9 μm) is wider than the cell (7.5 μm), so yes — barely. (c) Cell radius $r = 3.75 \times 10^{-6}$ m. Volume per cell: $V = \frac{4}{3}\pi (3.75 \times 10^{-6})^3 = \frac{4}{3}\pi \cdot 52.73 \times 10^{-18} \approx 2.21 \times 10^{-16}$ m³. Total: $5 \times 10^{12} \times 2.21 \times 10^{-16} \approx 1.1 \times 10^{-3}$ m³ — about **1.1 litres**, which matches real blood volume reasonably.

**Index laws investigation.** (a) Show that $\dfrac{x^5}{x^5} = 1$ and use this to explain why $x^0 = 1$. (b) Show that $\dfrac{x^3}{x^5} = \dfrac{1}{x^2}$ and use this to explain why $x^{-2} = \dfrac{1}{x^2}$. (c) Use the rule $(x^{1/2})^2 = x^1$ to explain why $x^{1/2} = \sqrt{x}$.

Answer

See working.

(a) By the division law: $\frac{x^5}{x^5} = x^{5-5} = x^0$. But $\frac{x^5}{x^5} = 1$ (any non-zero quantity divided by itself). So $x^0 = 1$. (b) Using the division law: $\frac{x^3}{x^5} = x^{3-5} = x^{-2}$. Directly: $\frac{x^3}{x^5} = \frac{1}{x^{5-3}} = \frac{1}{x^2}$. Combining: $x^{-2} = \frac{1}{x^2}$. This justifies the negative index rule. (c) Using the power rule: $(x^{1/2})^2 = x^{1/2 \cdot 2} = x^1 = x$. So $x^{1/2}$ is the number which squared gives $x$, i.e. the square root: $x^{1/2} = \sqrt{x}$.

**Population growth.** A town's population in 2020 is $4.5 \times 10^4$. It grows at 3% per year. (a) Write the population $P_n$ after $n$ years in the form $P_n = a \cdot b^n$. (b) Predict the population in 2030 (in standard form to 3 s.f.). (c) After how many years does the population first exceed $7 \times 10^4$?

Answer

(a) $P_n = 4.5 \times 10^4 \cdot 1.03^n$ (b) $\approx 6.05 \times 10^4$ (c) 15 years

(a) Compound growth multiplier 1.03 per year. $P_n = 4.5 \times 10^4 \cdot 1.03^n$. (b) After 10 years: $P_{10} = 4.5 \times 10^4 \cdot 1.03^{10} = 4.5 \times 10^4 \cdot 1.3439 \approx 6.05 \times 10^4$. (c) Solve $4.5 \times 10^4 \cdot 1.03^n > 7 \times 10^4$, i.e. $1.03^n > \frac{7}{4.5} = 1.556$. Test: $1.03^{14} \approx 1.513$ (no); $1.03^{15} \approx 1.558$ (yes). So after **15 years** (i.e. in 2035).

**Comparing magnitudes.** Place these in order from smallest to largest: $3 \times 10^{-2}, \quad 0.4, \quad 1.5 \times 10^{-1}, \quad \dfrac{1}{50}, \quad 6 \times 10^{-3}$

Answer

$6 \times 10^{-3} < \dfrac{1}{50} < 3 \times 10^{-2} < 1.5 \times 10^{-1} < 0.4$

Convert each to standard form or decimal: - $3 \times 10^{-2} = 0.03$ - $0.4 = 0.4$ - $1.5 \times 10^{-1} = 0.15$ - $\frac{1}{50} = 0.02$ - $6 \times 10^{-3} = 0.006$ Ordered: $0.006 < 0.02 < 0.03 < 0.15 < 0.4$, i.e. $6 \times 10^{-3} < \frac{1}{50} < 3 \times 10^{-2} < 1.5 \times 10^{-1} < 0.4$.

**Exponential decay.** A radioactive substance has a half-life of 5 days — every 5 days, half of it decays. A sample initially has mass 80 grams. (a) How much remains after 5, 10, 15, 20 days? (b) Write the mass $M$ after $5n$ days as a power expression. (c) After how many days is less than 1 gram remaining?

Answer

(a) 40, 20, 10, 5 (b) $M = 80 \cdot \left(\tfrac{1}{2}\right)^n$ (c) After 35 days (mass 0.625 g)

(a) Each 5-day period halves: 80 → 40 → 20 → 10 → 5. (b) $M = 80 \cdot \left(\frac{1}{2}\right)^n = \frac{80}{2^n}$ after $n$ half-lives ($= 5n$ days). (c) Need $\frac{80}{2^n} < 1$, i.e. $2^n > 80$. $2^6 = 64 < 80 < 128 = 2^7$. So $n = 7$ half-lives = **35 days**: $M = \frac{80}{128} = 0.625$ g.

**Surd/fractional index practice.** (a) Write $\sqrt{x^6}$ using indices and simplify. (b) Write $\sqrt[3]{x^9}$ in simplest index form. (c) Simplify $\dfrac{\sqrt{x^5}}{x^{1/2}}$. (d) Hence solve $\sqrt{x} = x^{1/2} = 9$.

Answer

(a) $x^3$ (b) $x^3$ (c) $x^2$ (d) $x = 81$

(a) $\sqrt{x^6} = (x^6)^{1/2} = x^{6/2} = x^3$. (b) $\sqrt[3]{x^9} = (x^9)^{1/3} = x^{9/3} = x^3$. (c) $\frac{\sqrt{x^5}}{x^{1/2}} = \frac{x^{5/2}}{x^{1/2}} = x^{5/2 - 1/2} = x^2$. (d) $x^{1/2} = 9 \Rightarrow x = 9^2 = 81$.

**The very large and very small.** A grain of sand has a typical mass of $4 \times 10^{-5}$ kg. The Earth has a mass of approximately $6 \times 10^{24}$ kg. (a) How many grains of sand would it take to equal the mass of the Earth? (b) The diameter of an atom is approximately $1 \times 10^{-10}$ m. The diameter of the Earth is approximately $1.3 \times 10^7$ m. How many atoms would span the Earth's diameter?

Answer

(a) $1.5 \times 10^{29}$ grains (b) $1.3 \times 10^{17}$ atoms

(a) Number of grains = $\frac{6 \times 10^{24}}{4 \times 10^{-5}} = \frac{6}{4} \times 10^{24-(-5)} = 1.5 \times 10^{29}$. (b) Number = $\frac{1.3 \times 10^7}{1 \times 10^{-10}} = 1.3 \times 10^{17}$.

**Common errors.** A student writes the following. Find and correct any errors. (a) $(x^3)^2 = x^5$ (b) $x^3 + x^3 = x^6$ (c) $x^0 = 0$ (d) $(2x)^3 = 2x^3$

Answer

All four are wrong. See working for corrections.

(a) **Wrong.** Power of a power *multiplies* indices: $(x^3)^2 = x^{3 \times 2} = x^6$. (b) **Wrong.** Adding *like terms* — the indices don't combine: $x^3 + x^3 = 2x^3$. (c) **Wrong.** Any non-zero number to the power 0 is 1: $x^0 = 1$ (for $x \neq 0$). (d) **Wrong.** Power applies to *both* the coefficient and the variable: $(2x)^3 = 2^3 \cdot x^3 = 8 x^3$.

**Simplify and evaluate.** Let $a = 2$ and $b = 3$. (a) Calculate $a^{-1} + b^{-1}$, giving your answer as a fraction. (b) Calculate $(a + b)^{-1}$. (c) Comment on whether $(a + b)^{-1} = a^{-1} + b^{-1}$ in general.

Answer

(a) $\dfrac{5}{6}$ (b) $\dfrac{1}{5}$ (c) Not in general

(a) $a^{-1} + b^{-1} = \frac{1}{2} + \frac{1}{3} = \frac{3}{6} + \frac{2}{6} = \frac{5}{6}$. (b) $(a + b)^{-1} = (2 + 3)^{-1} = 5^{-1} = \frac{1}{5}$. (c) $\frac{5}{6} \neq \frac{1}{5}$, so $(a+b)^{-1} \neq a^{-1} + b^{-1}$ in general. This is a common misconception: the negative index does **not** distribute over a sum. The general rule for fractions: $\frac{1}{a} + \frac{1}{b} = \frac{a+b}{ab}$, not $\frac{1}{a+b}$.

**Index equation with manipulation.** Solve for $x$: $$2^{x+3} = 4^{x-1}.$$

Answer

$x = 5$

Write both sides with base 2. RHS: $4^{x-1} = (2^2)^{x-1} = 2^{2(x-1)} = 2^{2x-2}$. Equation: $2^{x+3} = 2^{2x-2}$. Same base, so indices equal: $x + 3 = 2x - 2$. Rearranging: $5 = x$. So **$x = 5$**. Check: LHS = $2^8 = 256$; RHS = $4^4 = 256$ ✓.

**Light-years.** A light-year is the distance light travels in one year. Light travels at $3 \times 10^8$ m/s. (a) Show that 1 light-year is approximately $9.46 \times 10^{15}$ m (use 1 year = $3.154 \times 10^7$ s). (b) The nearest star (Proxima Centauri) is approximately 4.25 light-years away. How far is this in km, in standard form (3 s.f.)? (c) A spacecraft travels at $30000$ m/s. How many years would it take to reach Proxima Centauri?

Answer

(a) See working (b) $4.02 \times 10^{13}$ km (c) $\approx 42{,}500$ years

(a) Distance = speed × time = $(3 \times 10^8) \times (3.154 \times 10^7) = 9.462 \times 10^{15}$ m ≈ **$9.46 \times 10^{15}$ m** ✓. (b) Distance to Proxima Centauri = $4.25 \times 9.46 \times 10^{15}$ m = $40.205 \times 10^{15}$ m = $4.02 \times 10^{16}$ m. Convert to km (÷ 1000): $4.02 \times 10^{13}$ km. (c) Time = $\frac{4.02 \times 10^{16} \text{ m}}{3 \times 10^4 \text{ m/s}} = 1.34 \times 10^{12}$ s. In years: $\frac{1.34 \times 10^{12}}{3.154 \times 10^7} \approx 4.25 \times 10^4 = \mathbf{42{,}500}$ years.