Indices: Grade 10 Math – Science Playground

Charcoal from an ancient fireplace has only $\dfrac{1}{16}$ of the carbon-14 that a living tree has. How old is it?

Carbon-14 halves every 5,730 years or so. After n of these half-lives, $\left(\dfrac12\right)^n$ of it is left, so the question is an equation with the unknown in the index: $\left(\dfrac12\right)^n = \dfrac{1}{16}$ . The answer is 4 half-lives, about 23,000 years.

By the end of this unit you will solve equations like this one, and see what powers such as $2^0$ , $2^{-1}$ and $8^{1/3}$ mean.

(The half-life: Wikipedia, “Radiocarbon dating”.)

Move the age of a dead plant or a piece of charcoal. The curve shows how much of its carbon-14 is left; the dots mark whole half-lives.

Age: 5,730 years

Half-lives: 1.00
Carbon-14 left: 50%
As a power: (1/2)¹

A plant that has just died has all the carbon-14 a living plant has: call that 100%.

After 1 half-life the carbon-14 has been halved once: (1/2)¹ = 1/2 of it is left.

Between whole half-lives the index is a fraction, and (1/2)^1.00 still has a value: the curve is smooth, not steps.

For the SEE

The question: Algebra is 15 marks in three questions. In CDC’s model paper one question of 5 marks has two parts: a. solve an exponential equation (3 marks, application), such as $25^x - 6 \times 5^x + 5 = 0$ , b. simplify an algebraic fraction (2 marks, Unit 8).
The rules it needs:

If $a^x = a^y$ (with a > 0, a ≠ 1), then x = y: make the bases equal.
$a^{x + 1} = a \times a^x$ and $a^{x - 1} = \dfrac{a^x}{a}$ : take $a^x$ out as a common factor.
$25^x = (5^x)^2$ : with $5^x = a$ the equation becomes a quadratic in a.
$a^0 = 1$ , $a^{-n} = \dfrac{1}{a^n}$ , $a^{1/n} = \sqrt[n]{a}$ ; a power of a positive base is never 0 or negative.

How the SEE asks it has a full question marked part by part; Exam practice has every exercise question to mark yourself.

9.0 Review

In $2^5 = 32$ , 2 is the base, 5 is the index (also called the power or exponent), and 32 is the value. A positive whole-number index counts how many times the base is multiplied: $2^5 = 2 \times 2 \times 2 \times 2 \times 2$ .

Zero, negative and fractional indices

“Multiplied by itself zero times” or “minus three times” makes no sense as counting. But look at the powers of 2 in a row: each step to the right multiplies by 2, so each step to the left divides by 2. Keep going left past $2^1$ :

$\cdots \quad 2^{-2} = \frac14 \quad 2^{-1} = \frac12 \quad 2^0 = 1 \quad 2^1 = 2 \quad 2^2 = 4 \quad \cdots$

These are the only values that keep the pattern, and they also keep the rule $a^m \times a^n = a^{m + n}$ true: $2^3 \times 2^0 = 2^3$ needs $2^0 = 1$ , and $2^3 \times 2^{-3} = 2^0 = 1$ needs $2^{-3} = \dfrac18$ .

A half step works the same way. Two half steps make one step, so $2^{1/2} \times 2^{1/2} = 2^1 = 2$ : $2^{1/2}$ is the number that, multiplied by itself, gives 2, which is $\sqrt2$ .

Choose a base and move the index, in half steps, from −3 to 3. Each step to the right multiplies by the base; watch what happens at 0, below 0 and halfway between.

Base:

Index: −2

Power: 2⁻²
Exactly: 1/4
About: 0.25

Each step to the right multiplies by 2, so 2^-2 is -2 2s multiplied together.

One step left of 2¹ divides 2 by 2: so 2⁰ = 1, and the same happens with every base.

Each step to the left divides by 2, past 1 as well: 2⁻² = 1 ÷ 2². The values get small but never reach 0.

Half a step: two half steps make one step, ×2, so a half step multiplies by √2 (√2 × √2 = 2). That is why 2^1/2 = √2.

The laws of indices, from Grade 9, for a > 0:

Law	Example
$a^m \times a^n = a^{m + n}$	$3^4 \times 3^2 = 3^6$
$a^m \div a^n = a^{m - n}$	$5^7 \div 5^5 = 5^2$
$(a^m)^n = a^{mn}$	$(2^3)^2 = 2^6$
$(ab)^n = a^n b^n$	$6^x = 2^x \times 3^x$
$a^0 = 1$	$7^0 = 1$
$a^{-n} = \dfrac{1}{a^n}$	$4^{-2} = \dfrac{1}{16}$
$a^{1/n} = \sqrt[n]{a}$	$8^{1/3} = \sqrt[3]{8} = 2$
$a^{m/n} = \left(\sqrt[n]{a}\right)^m$	$8^{2/3} = 2^2 = 4$

What is $5^0$ ?

Pick an answer to see why.

What is $2^{-3}$ ?

Pick an answer to see why.

Which index gives the value?

Fill in each table, then find the value of x that satisfies the equation.

a. $2^x = 2$

x	−3	−2	−1	0	1	2	3
$2^x$	$2^{-3} = \frac18$	…	…	…	…	…	…

b. $5^{x + 1} = 125$

x	−3	−2	−1	0	1	2	3
$5^{x + 1}$	$5^{-2} = \frac{1}{25}$	…	…	…	…	…	…

c. $3^x = \dfrac19$

x	−3	−2	−1	0	1	2	3
$3^x$	$3^{-3} = \frac{1}{27}$	…	…	…	…	…	…

9.1 Exponential Equations

An equation with the unknown in the index, such as $2^x = 4$ , is an exponential equation. Trying x = 0, ±1, ±2, … one by one works for easy ones, but not for $3^x + \dfrac{1}{3^x} = 3\dfrac13$ . There is a better way.

The idea behind a and b is a rule:

If $a^x = a^y$ , where a > 0 and a ≠ 1, then x = y.

It works because the powers of such a base never repeat a value: $2^x$ only ever grows as x grows. (It fails for the base 1: $1^3 = 1^7$ , but 3 ≠ 7.) So the plan is always the same: write both sides as powers of the same base, then equate the indices.

Three kinds of exponential equation

One power on each side: make the bases equal (Examples 1 and 2).
Several powers of the same base added: take a power out as a common factor; the smallest keeps the numbers whole (Examples 3 and 4).
A power and its square, or a power and its reciprocal: let the power be a and solve a quadratic (Examples 5 and 6).

Worked examples

In solving $9^x - 4 \times 3^x + 3 = 0$ , what should you let a be?

Pick an answer to see why.

Proving a relation with cube roots

Some questions give x as a sum of cube roots and ask you to prove a cubic relation. The trick is to cube both sides, using

$(p + q)^3 = p^3 + q^3 + 3pq(p + q)$

The p + q at the end is x again, so x comes back into the equation.

Back to the charcoal

The charcoal has $\dfrac{1}{16}$ of a living tree’s carbon-14. After n half-lives, $\left(\dfrac12\right)^n$ is left:

$\begin{aligned} \left(\frac12\right)^n &= \frac{1}{16} \\ 2^{-n} &= 2^{-4} \\ n &= 4 \end{aligned}$

Four half-lives: 4 × 5730 = 22920, about 23,000 years. That is a fire from the last Ice Age.

The index needn’t be whole. A wooden bowl with $\dfrac{1}{\sqrt2}$ of the carbon-14 (about 71%) gives $2^{-n} = 2^{-1/2}$ , so $n = \dfrac12$ : half a half-life, about 2,900 years.

How the SEE asks it

In the SEE’s Compulsory Mathematics paper, Algebra carries 15 marks in three questions (CDC’s specification grid). In CDC’s model paper this unit shares a 5-mark question with Unit 8:

Part	Level	Marks	What it asks
a.	application	3	solve an exponential equation that becomes a quadratic
b.	understanding	2	simplify algebraic fractions (Unit 8)

The 3 marks of part a are usually one for turning it into a quadratic, one for its roots and one for the values of x. Here is a question in that shape.

Exercise 9.1

Question 1

Fill in the tables.

x	−3	−2	−1	0	1	2	3
$7^x$	…	…	…	…	…	…	…

Answer

x	−3	−2	−1	0	1	2	3
$5^{-x}$	…	…	…	…	…	…	…

Answer

Question 2

Solve, and check your answer.

$3^x = 9$

Answer

$5^{x - 1} = 25$

Answer

$\dfrac{1}{5^{2x - 4}} = 125$

Answer

$4^{x - 2} = 0.125$

Answer

$\left(\dfrac35\right)^x = \left(1\dfrac23\right)^3$

Answer

$2^x \times 3^{x + 1} = 18$

Answer

Question 3

Solve.

$4^{\frac{1 - x}{1 + x}} = 4^{\frac13}$

Answer

$\sqrt[2x + 4]{4^{x + 8}} = \sqrt[6]{128}$

Hint

Answer

$2^{x + 1} + 2^{x + 2} + 2^{x + 3} = 448$

Answer

$3^{x + 1} - 3^x = 162$

Answer

$4^{x + 1} - 8 \times 4^{x - 1} = 32$

Answer

$4 \times 3^{x + 1} - 3^{x + 2} - 3^{x - 1} = 72$

Answer

$3^{x + 2} + 3^{x + 1} + 2 \times 3^x = 126$

Answer

$2^x + 3^{x - 2} = 3^x - 2^{x + 1}$

Hint

Answer

$8^{x - 1} - 2^{3x - 2} + 8 = 0$

Answer

$\left(\dfrac14\right)^{2 - \sqrt{5x + 1}} = 4 \times 2^{\sqrt{5x + 1}}$

Hint

Answer

Question 4

Solve.

$5^x + \dfrac{1}{5^x} = 5\dfrac15$

Answer

$7^x + \dfrac{1}{7^x} = 7\dfrac17$

Answer

$9^x + \dfrac{1}{9^x} = 9\dfrac19$

Answer

$4^x + \dfrac{1}{4^x} = 16\dfrac{1}{16}$

Answer

$5^x + 5^{-x} = 25\dfrac{1}{25}$

Answer

$81 \times 3^x + 3^{-x} = 30$

Answer

Question 5

Solve.

$4 \times 3^{x + 1} - 9^x = 27$

Answer

$3 \times 2^{p + 1} - 4^p = 8$

Answer

$5^{2x} - 6 \times 5^{x + 1} + 125 = 0$

Answer

$2^{x - 2} + 2^{3 - x} = 3$

Answer

$5^{x + 1} + 5^{2 - x} = 126$

Answer

$3^{2y} - 4 \times 3^y + 3 = 0$

Answer

Question 6

Solve $16^x - 5 \times 4^{x + 1} + 64 = 0$ . Show that each value of x satisfies $5^x + \dfrac{125}{5^x} = 30$ .

Answer

Question 7

If $x = 3^{1/3} + 3^{-1/3}$ , prove that $3x(x^2 - 3) = 10$ .

Answer

If $x = 2^{1/3} - 2^{-1/3}$ , prove that $2x^3 + 6x - 3 = 0$ .

Answer

Mixed exercise

These questions mix the four algebra units: Unit 6 (sequences and series), Unit 7 (quadratic equations), Unit 8 (algebraic fractions) and this unit, as the SEE’s Algebra questions do.

Question 1

Two employees of a bookshop, Nepal Pustak Pasal, earned this commission in five months.

Month	Employee A	Employee B
Baishakh	Rs 5,000	Rs 2,000
Jestha	Rs 6,000	Rs 3,000
Ashadh	Rs 7,000	Rs 4,500
Shrawan	Rs 8,000	Rs 6,750
Bhadra	Rs 9,000	Rs 10,125

Whose commission forms an arithmetic sequence? Give a reason.

Answer

Find the mean of each employee’s Baishakh and Ashadh commission: the arithmetic mean for A and the geometric mean for B.

Answer

Using the formulas for series, find the difference between their total commissions for the five months.

Answer

Question 2

Bishal borrowed Rs 45,000 from his friend Sunil, to repay in 6 installments, each Rs 1,000 more than the one before. Sita borrowed Rs 63,000 from her friend Om Kumari, to repay in 6 installments, each double the one before.

How much does each of them pay in the first installment?

Answer

Find the difference between their first installments, and between their last installments.

Answer

In which installment do they pay the same amount?

Answer

Question 3

Two birds came to a community forest on the first day. Each day after, twice as many birds came as the day before.

How many birds come on the tenth day?

Answer

How many birds come in the ten days altogether?

Answer

Question 4

From Baishakh 1 to Baishakh 7, Nabraj’s father put money in Nabraj’s piggy bank, each day double the day before. After the seventh day it held Rs 635.

How much did he put in on the first day?

Answer

How much did he put in on the seventh day?

Answer

Question 5

Sunil’s father puts money aside on each of Sunil’s birthdays: Rs 500 on the first, Rs 1,000 on the second, Rs 1,500 on the third, and Rs 500 more each year.

How much is put aside on his 16th birthday?

Answer

How much has been put aside in all by his 16th birthday?

Answer

On which birthday does the total first reach Rs 1 lakh? Give a reason.

Answer

Question 6

Harisaran has two rectangular fields, each of area 360 m². In the first the length is 2 m more than the breadth; in the second it is 9 m more. He wants to fence both with barbed wire.

Find the length and breadth of each field.

Answer

Do the two fields need the same length of wire?

Answer

At Rs 10 a meter, which field costs more to fence once round, and by how much?

Answer

Question 7

Ram is 30 years old and his wife Sita is 25.

What were their ages x years ago?

Answer

x years ago the product of their ages was 500. Find x.

Answer

In how many years will the sum of their ages be 99?

Answer

Question 8

The first term of an arithmetic sequence is 2, and the sum of the next five terms is four times the sum of the first five.

Find the common difference.

Answer

Show that $t_{20} = -112$ .

Answer

Find the sum of the first five terms.

Answer

Question 9

A footpath of the same width runs all round a 16 m by 12 m grassland. The grassland and the footpath together cover 320 m².

Figure 9.1 The grassland and the footpath round it.

Taking the width of the footpath as x m, write an equation.

Answer

Find the width of the footpath.

Answer

Question 10

In a rectangular field the longer side is 40 m more than the shorter side, and the diagonal is 40 m more than the longer side.

Figure 9.2 The field and its diagonal.

Taking the shorter side as x m, write an equation.

Answer

Find the shorter side, the longer side and the diagonal.

Answer

How much does it cost to fence the field four times round with barbed wire at Rs 15 a meter?

Answer

How many plots of 20 m by 15 m can the field be divided into?

Answer

Question 11

Ramesh is 30 years old and his sister Sita is 25.

What were their ages x years ago?

Answer

x years ago the product of their ages was 644. Find x.

Answer

In how many years will the product of their ages be 864?

Answer

Question 12

Two cars leave a crossroads at the same time, one going north and the other west. When the northbound car has gone 24 miles, the distance between the cars is 4 miles more than three times the distance the westbound car has gone.

Figure 9.3 The cars’ distances, in miles, drawn to scale with the answer: north up the page, west to the left.

Write an equation.

Answer

How far has the westbound car gone?

Answer

How far apart are the cars?

Answer

Question 13

A bus travels 90 km at a steady speed. Had it gone 15 km/h faster, the journey would have taken 30 minutes less.

Taking the speed as x km/h, write an equation.

Answer

Find the bus’s speed.

Answer

How long did the journey take?

Answer

Question 14

Simplify.

$\dfrac{1}{a - b} - \dfrac{2b}{a^2 - b^2}$

Answer

$\dfrac{a - 4}{a^2 - 4a + 16} + \dfrac{a + 4}{a^2 + 4a + 16} + \dfrac{128}{a^4 + 16a^2 + 256}$

Hint

Answer

$\dfrac{2a - 6}{a^2 - 9a + 20} - \dfrac{a - 1}{a^2 - 7a + 12} - \dfrac{a - 2}{a^2 - 8a + 15}$

Answer

$\dfrac{a + b}{2ab}(a + b - c) + \dfrac{b + c}{2bc}(b + c - a) + \dfrac{c + a}{2ca}(c + a - b)$

Answer

Question 15

Solve.

$3^{x + 2} + 3^{2 - x} = 82$

Answer

$\dfrac{3^{2x} + 1}{3^x} = \dfrac{82}{9}$

Answer

Question 16

Prove each relation.

If $x = 1 + 2^{1/3} + 2^{2/3}$ , then $x(x^2 - 3x - 3) = 1$ .

Hint

Answer

If $x = 3 + 3^{1/3} + 3^{2/3}$ , then $x(x^2 - 9x + 18) = 12$ .

Answer

If $x = 2 - 2^{1/3} + 2^{2/3}$ , then $x(x^2 - 6x + 18) = 22$ .

Answer

Worksheet to print (PDF, 4 pages) Answer key for teachers

Harder worksheet (PDF, 3 pages) Answer key for teachers

Quiz yourself on this unit →

Practice every question in the book, and keep track of the ones you miss →

Where these facts come from

Carbon-14’s half-life, about 5,730 years (measured more recently as 5,700 ± 30), the limit of about 50,000 years, and Libby’s work at Chicago and his 1960 Nobel Prize (the opening, Back to the charcoal, the story): Wikipedia, “Radiocarbon dating”, read October 2026.
A0’s area of 1 m², the √2 ratio of the sides, each size half the one before, and A4’s 210 mm by 297 mm (fact box): Wikipedia, “ISO 216”, read October 2026.
The SEE’s Algebra marks and the model question (For the SEE, How the SEE asks it): CDC, Specification Grid, Compulsory Mathematics, Grades 9–10 (2078), the grid table and the Grade 10 model question paper, question 10.

A companion to your textbook, not a replacement for it: study from the book your school uses, and where the two differ, follow your teacher. Found a mistake? Please tell us.