Sec 3 · G3 (Express)

Indices, compound interest and standard form

Practise indices, compound interest and standard form with step-by-step help. Free account.

1. Laws of indices

Multiply: add the indices

\displaystyle a^{m} \times a^{n} = a^{m + n}

e.g.

\displaystyle 2a^{5} \times 3a^{3}

\displaystyle = 6a^{8}

Divide: subtract the indices

\displaystyle a^{m} \div a^{n} = a^{m - n}

e.g.

\displaystyle \dfrac{12a^{7}}{4a^{3}}

\displaystyle = 3a^{4}

Power of a power: multiply the indices

\displaystyle (a^{m})^{n} = a^{mn}

e.g.

\displaystyle (2a^{3})^{4}

\displaystyle = 16a^{12}

Power of a product

\displaystyle (ab)^{n} = a^{n}b^{n}

e.g.

\displaystyle (3ab^{2})^{2}

\displaystyle = 9a^{2}b^{4}

Power of a quotient

\displaystyle \left(\dfrac{a}{b}\right)^{n} = \dfrac{a^{n}}{b^{n}}

e.g.

\displaystyle \left(\dfrac{a^{2}}{3}\right)^{3}

\displaystyle = \dfrac{a^{6}}{27}

📖 Worked example

Simplify $\dfrac{6a^{8}}{3a^{2}}$ .

Step 1 of 2

Simplify the numbers, then the power of each letter

Numbers:

\dfrac{6}{3} = 2

a

: subtract the indices,

8 - 2 = 6

📖 Worked example

Simplify $(2a^{3})^{4}$ .

Step 1 of 2

Simplify the number in front, then the power of each letter

Numbers:

2^{4} = 16

a

: multiply the indices,

3 \times 4 = 12

Practise this →

2. Zero, negative and fractional indices

Zero index

\displaystyle a^{0} = 1, \quad a \neq 0

e.g.

\displaystyle 7^{0} = 1

Negative index

\displaystyle a^{-n} = \dfrac{1}{a^{n}}

e.g.

\displaystyle 2^{-3}

\displaystyle = \dfrac{1}{2^{3}}

\displaystyle = \dfrac{1}{8}

Negative index of a fraction: flip it

\displaystyle \left(\dfrac{a}{b}\right)^{-n} = \left(\dfrac{b}{a}\right)^{n}

e.g.

\displaystyle \left(\dfrac{2}{3}\right)^{-2}

\displaystyle = \left(\dfrac{3}{2}\right)^{2}

\displaystyle = \dfrac{9}{4}

Root

\displaystyle a^{\frac{1}{n}} = \sqrt[n]{a}

e.g.

\displaystyle 64^{\frac{1}{3}}

\displaystyle = \sqrt[3]{64}

\displaystyle = 4

Fractional index: root first, then power

\displaystyle a^{\frac{m}{n}} = \left(\sqrt[n]{a}\right)^{m}

e.g.

\displaystyle 27^{\frac{2}{3}}

\displaystyle = \left(\sqrt[3]{27}\right)^{2}

\displaystyle = 3^{2}

\displaystyle = 9

Negative fractional index

\displaystyle a^{-\frac{m}{n}} = \dfrac{1}{\left(\sqrt[n]{a}\right)^{m}}

e.g.

\displaystyle 16^{-\frac{3}{4}}

\displaystyle = \dfrac{1}{\left(\sqrt[4]{16}\right)^{3}}

\displaystyle = \dfrac{1}{2^{3}}

\displaystyle = \dfrac{1}{8}

📖 Worked example

Evaluate $16^{-\frac{3}{4}}$ .

Step 1 of 4

Rewrite with a positive index

16^{-\frac{3}{4}} = \left(\dfrac{1}{16}\right)^{\frac{3}{4}}

📖 Worked example

Simplify $(8a^{3}b^{-6})^{-\frac{2}{3}}$ , giving your answer with positive indices.

Step 1 of 2

Apply the power to every part of the bracket

Numbers:

8^{-\frac{2}{3}} = 2^{-2}

8^{-\frac{2}{3}} = \dfrac{1}{4}

a

3 \times \left(-\dfrac{2}{3}\right) = -2

b

-6 \times \left(-\dfrac{2}{3}\right) = 4

Practise this →

3. Index equations

Same base: the indices are equal

\displaystyle a^{m} = a^{n} \;\Rightarrow\; m = n

e.g.

\displaystyle 4^{x - 1} = 64

\displaystyle 2^{2(x - 1)} = 2^{6}

\displaystyle 2(x - 1) = 6

\displaystyle x = 4

A fraction is a negative power

\displaystyle \dfrac{1}{a^{n}} = a^{-n}

e.g.

\displaystyle 8^{x + 3} = \dfrac{1}{16}

\displaystyle 2^{3(x + 3)} = 2^{-4}

\displaystyle 3(x + 3) = -4

\displaystyle x = -\dfrac{13}{3}

📖 Worked example

Solve $4^{x - 1} = 64$ .

Step 1 of 5

Write both sides as powers of the same number

2^{2(x - 1)} = 2^{6}

Equation	First move	Then
$3^{x + 2} - 3^{x} = 216$	take out $3^{x}$ : $3^{x}(3^{2} - 1) = 216$	$3^{x} = 27$ , $x = 3$
$2^{x} + 2^{x} = 64$	$2 \times 2^{x} = 2^{x + 1}$	$x + 1 = 6$ , $x = 5$
$2^{x} \times 5^{x} = 1000$	same index: $10^{x}$	$x = 3$
$3^{x} \times 9^{3} = \dfrac{1}{81^{x}}$	$3^{x + 6} = 3^{-4x}$	$x = -\dfrac{6}{5}$

In terms of a, where a = 3ˣ

\displaystyle 9^{x + 1} = 9(3^{x})^{2}

$9^{x + 1} = 3^{2x} \times 3^{2}$
so $9^{x + 1} = 9a^{2}$

📖 Worked example

Solve the equation $3^{x + 2} - 3^{x} = 216$ .

Step 1 of 3

Take out the lowest power, 3^x, as a common factor

3^{x}\left(3^{2} - 1\right) = 216

3^{x} \times 8 = 216

📖 Worked example

Given that $a = 3^{x}$ , express $9^{x + 1}$ in terms of $a$ .

Step 1 of 3

Write 9 as a power of 3

9 = 3^{2}

Practise this →

4. Compound interest

Compound interest

\displaystyle A = P\left(1 + \dfrac{r}{100}\right)^{n}

Interest earned

\displaystyle \text{Interest} = A - P

Depreciation (losing value)

\displaystyle A = P\left(1 - \dfrac{r}{100}\right)^{n}

Symbol	Meaning	Example
$A$	total amount at the end	$5624.32
$P$	amount invested (principal)	$5000
$r$	interest rate per annum, %	$4\%$ p.a.
$n$	number of years	$3$ years

Compounded	Rate per period	Number of periods
yearly	$r$	$n$
half-yearly	$\dfrac{r}{2}$	$2n$
quarterly	$\dfrac{r}{4}$	$4n$
monthly	$\dfrac{r}{12}$	$12n$

📖 Worked example

Mrs Tan invests $5000 at 3% per annum compound interest, compounded half-yearly. Find the total amount after 3 years, correct to the nearest cent.

Step 1 of 3

Use the compound interest formula, with the rate and the number of periods for each compounding period

A = P\left(1 + \dfrac{r}{100}\right)^n

A = 5000\left(1 + \dfrac{3}{100 \times 2}\right)^{6}

📖 Worked example

The value of a car bought for $25 000 depreciates at 15% per annum. Find its value after 2 years, correct to the nearest cent.

Step 1 of 2

Use the compound interest formula

A = P\left(1 - \dfrac{r}{100}\right)^n

A = 25000\left(1 - \dfrac{15}{100}\right)^{2}

📖 Worked example

Priya invests some money at 3% per annum compound interest, compounded annually. After 3 years the investment is worth $8800. Find the amount invested, correct to the nearest cent.

Step 1 of 2

Write the formula with the unknown amount

8800 = P\left(1 + \dfrac{3}{100}\right)^{3}

Question	Method
a different rate in the first year	$P\left(1 + \dfrac{r_1}{100}\right)\left(1 + \dfrac{r_2}{100}\right)^{n}$
least whole number of years to pass a target	try $n = 1, 2, 3, \ldots$ : the first year past the target

📖 Worked example

Mrs Tan invests $5000 at 2.5% per annum compound interest, compounded yearly. Find the least number of whole years after which the investment is worth more than $6300.

Step 1 of 3

The value after n years

5000 \times 1.025^{n}

Practise this →

5. Standard form

Standard form

\displaystyle A \times 10^{n}, \quad 1 \leq A < 10

$n$ is an integer

Number	Standard form	Point moves
$3\,450\,000$	$3.45 \times 10^{6}$	6 places left
$0.000\,345$	$3.45 \times 10^{-4}$	4 places right
$45\,000$	$4.5 \times 10^{4}$	4 places left
$0.07$	$7 \times 10^{-2}$	2 places right

Multiply: multiply the numbers, add the powers

\displaystyle (a \times 10^{m}) \times (b \times 10^{n}) = ab \times 10^{m + n}

e.g.

\displaystyle (4 \times 10^{3}) \times (5 \times 10^{2})

\displaystyle = 20 \times 10^{5}

\displaystyle = 2 \times 10^{6}

Divide: divide the numbers, subtract the powers

\displaystyle \dfrac{a \times 10^{m}}{b \times 10^{n}} = \dfrac{a}{b} \times 10^{m - n}

e.g.

\displaystyle (9 \times 10^{5}) \div (3 \times 10^{2})

\displaystyle = 3 \times 10^{3}

Add or subtract: same power of 10 first

\displaystyle a \times 10^{n} + b \times 10^{n} = (a + b) \times 10^{n}

e.g.

\displaystyle 3.2 \times 10^{4} + 6.4 \times 10^{3}

\displaystyle = 3.2 \times 10^{4} + 0.64 \times 10^{4}

\displaystyle = 3.84 \times 10^{4}

📖 Worked example

Express $0.000\,345$ in standard form.

Step 1 of 2

Move the decimal point to just after the first non-zero digit

Number of places the point moves:

4

📖 Worked example

Calculate $(3.2 \times 10^{4}) + (6.4 \times 10^{3})$ , giving your answer in standard form.

Step 1 of 2

Write both with the same power of 10

6.4 \times 10^{3} = 0.64 \times 10^{4}

Prefix	Symbol	Power of 10
giga	G	$10^{9}$
mega	M	$10^{6}$
kilo	k	$10^{3}$
milli	m	$10^{-3}$
micro	$\mu$	$10^{-6}$
nano	n	$10^{-9}$

A given power of 10 (not standard form)

\displaystyle 9.6 \times 10^{-13} = 0.96 \times 10^{-12}

Powers with letters

\displaystyle (4 \times 10^{m})(6 \times 10^{n}) = 2.4 \times 10^{m + n + 1}

$24 \times 10^{m + n}$ is not standard form: $24 = 2.4 \times 10^{1}$

📖 Worked example

1 nanometre $(\text{nm}) = 10^{-9}$ metres. Express $450$ $\text{nm}$ in metres, giving your answer in standard form.

Step 1 of 3

Replace the prefix nano by its power of 10

450\ \text{nm} = 450 \times 10^{-9}

metres

Practise this →

Try a question: Indices, compound interest and standard form

Free · no sign-up

Answer a real practice question. Get it wrong and MugCorner shows you where it went wrong, with hints and a step-by-step guide.

Practise indices, compound interest and standard form with step-by-step help

Endless questions from basic to exam level. Get one wrong and see where it went wrong. Free account.

Start free

Practice questions by skill

Sec 3 · G3 (Express)

Indices, compound interest and standard form

Practise indices, compound interest and standard form with step-by-step help. Free account.

Start free Worksheet (PDF)

1. Laws of indices

Multiply: add the indices

\displaystyle a^{m} \times a^{n} = a^{m + n}

e.g.

\displaystyle 2a^{5} \times 3a^{3}

\displaystyle = 6a^{8}

Divide: subtract the indices

\displaystyle a^{m} \div a^{n} = a^{m - n}

e.g.

\displaystyle \dfrac{12a^{7}}{4a^{3}}

\displaystyle = 3a^{4}

Power of a power: multiply the indices

\displaystyle (a^{m})^{n} = a^{mn}

e.g.

\displaystyle (2a^{3})^{4}

\displaystyle = 16a^{12}

Power of a product

\displaystyle (ab)^{n} = a^{n}b^{n}

e.g.

\displaystyle (3ab^{2})^{2}

\displaystyle = 9a^{2}b^{4}

Power of a quotient

\displaystyle \left(\dfrac{a}{b}\right)^{n} = \dfrac{a^{n}}{b^{n}}

e.g.

\displaystyle \left(\dfrac{a^{2}}{3}\right)^{3}

\displaystyle = \dfrac{a^{6}}{27}

📖 Worked example

Simplify $\dfrac{6a^{8}}{3a^{2}}$ .

Step 1 of 2

Simplify the numbers, then the power of each letter

Numbers:

\dfrac{6}{3} = 2

a

: subtract the indices,

8 - 2 = 6

📖 Worked example

Simplify $(2a^{3})^{4}$ .

Step 1 of 2

Simplify the number in front, then the power of each letter

Numbers:

2^{4} = 16

a

: multiply the indices,

3 \times 4 = 12

Practise this →

2. Zero, negative and fractional indices

Zero index

\displaystyle a^{0} = 1, \quad a \neq 0

e.g.

\displaystyle 7^{0} = 1

Negative index

\displaystyle a^{-n} = \dfrac{1}{a^{n}}

e.g.

\displaystyle 2^{-3}

\displaystyle = \dfrac{1}{2^{3}}

\displaystyle = \dfrac{1}{8}

Negative index of a fraction: flip it

\displaystyle \left(\dfrac{a}{b}\right)^{-n} = \left(\dfrac{b}{a}\right)^{n}

e.g.

\displaystyle \left(\dfrac{2}{3}\right)^{-2}

\displaystyle = \left(\dfrac{3}{2}\right)^{2}

\displaystyle = \dfrac{9}{4}

Root

\displaystyle a^{\frac{1}{n}} = \sqrt[n]{a}

e.g.

\displaystyle 64^{\frac{1}{3}}

\displaystyle = \sqrt[3]{64}

\displaystyle = 4

Fractional index: root first, then power

\displaystyle a^{\frac{m}{n}} = \left(\sqrt[n]{a}\right)^{m}

e.g.

\displaystyle 27^{\frac{2}{3}}

\displaystyle = \left(\sqrt[3]{27}\right)^{2}

\displaystyle = 3^{2}

\displaystyle = 9

Negative fractional index

\displaystyle a^{-\frac{m}{n}} = \dfrac{1}{\left(\sqrt[n]{a}\right)^{m}}

e.g.

\displaystyle 16^{-\frac{3}{4}}

\displaystyle = \dfrac{1}{\left(\sqrt[4]{16}\right)^{3}}

\displaystyle = \dfrac{1}{2^{3}}

\displaystyle = \dfrac{1}{8}

📖 Worked example

Evaluate $16^{-\frac{3}{4}}$ .

Step 1 of 4

Rewrite with a positive index

16^{-\frac{3}{4}} = \left(\dfrac{1}{16}\right)^{\frac{3}{4}}

📖 Worked example

Simplify $(8a^{3}b^{-6})^{-\frac{2}{3}}$ , giving your answer with positive indices.

Step 1 of 2

Apply the power to every part of the bracket

Numbers:

8^{-\frac{2}{3}} = 2^{-2}

8^{-\frac{2}{3}} = \dfrac{1}{4}

a

3 \times \left(-\dfrac{2}{3}\right) = -2

b

-6 \times \left(-\dfrac{2}{3}\right) = 4

Practise this →

3. Index equations

Same base: the indices are equal

\displaystyle a^{m} = a^{n} \;\Rightarrow\; m = n

e.g.

\displaystyle 4^{x - 1} = 64

\displaystyle 2^{2(x - 1)} = 2^{6}

\displaystyle 2(x - 1) = 6

\displaystyle x = 4

A fraction is a negative power

\displaystyle \dfrac{1}{a^{n}} = a^{-n}

e.g.

\displaystyle 8^{x + 3} = \dfrac{1}{16}

\displaystyle 2^{3(x + 3)} = 2^{-4}

\displaystyle 3(x + 3) = -4

\displaystyle x = -\dfrac{13}{3}

📖 Worked example

Solve $4^{x - 1} = 64$ .

Step 1 of 5

Write both sides as powers of the same number

2^{2(x - 1)} = 2^{6}

Equation	First move	Then
$3^{x + 2} - 3^{x} = 216$	take out $3^{x}$ : $3^{x}(3^{2} - 1) = 216$	$3^{x} = 27$ , $x = 3$
$2^{x} + 2^{x} = 64$	$2 \times 2^{x} = 2^{x + 1}$	$x + 1 = 6$ , $x = 5$
$2^{x} \times 5^{x} = 1000$	same index: $10^{x}$	$x = 3$
$3^{x} \times 9^{3} = \dfrac{1}{81^{x}}$	$3^{x + 6} = 3^{-4x}$	$x = -\dfrac{6}{5}$

In terms of a, where a = 3ˣ

\displaystyle 9^{x + 1} = 9(3^{x})^{2}

$9^{x + 1} = 3^{2x} \times 3^{2}$
so $9^{x + 1} = 9a^{2}$

📖 Worked example

Solve the equation $3^{x + 2} - 3^{x} = 216$ .

Step 1 of 3

Take out the lowest power, 3^x, as a common factor

3^{x}\left(3^{2} - 1\right) = 216

3^{x} \times 8 = 216

📖 Worked example

Given that $a = 3^{x}$ , express $9^{x + 1}$ in terms of $a$ .

Step 1 of 3

Write 9 as a power of 3

9 = 3^{2}

Practise this →

4. Compound interest

Compound interest

\displaystyle A = P\left(1 + \dfrac{r}{100}\right)^{n}

Interest earned

\displaystyle \text{Interest} = A - P

Depreciation (losing value)

\displaystyle A = P\left(1 - \dfrac{r}{100}\right)^{n}

Symbol	Meaning	Example
$A$	total amount at the end	$5624.32
$P$	amount invested (principal)	$5000
$r$	interest rate per annum, %	$4\%$ p.a.
$n$	number of years	$3$ years

Compounded	Rate per period	Number of periods
yearly	$r$	$n$
half-yearly	$\dfrac{r}{2}$	$2n$
quarterly	$\dfrac{r}{4}$	$4n$
monthly	$\dfrac{r}{12}$	$12n$

📖 Worked example

Mrs Tan invests $5000 at 3% per annum compound interest, compounded half-yearly. Find the total amount after 3 years, correct to the nearest cent.

Step 1 of 3

Use the compound interest formula, with the rate and the number of periods for each compounding period

A = P\left(1 + \dfrac{r}{100}\right)^n

A = 5000\left(1 + \dfrac{3}{100 \times 2}\right)^{6}

📖 Worked example

The value of a car bought for $25 000 depreciates at 15% per annum. Find its value after 2 years, correct to the nearest cent.

Step 1 of 2

Use the compound interest formula

A = P\left(1 - \dfrac{r}{100}\right)^n

A = 25000\left(1 - \dfrac{15}{100}\right)^{2}

📖 Worked example

Priya invests some money at 3% per annum compound interest, compounded annually. After 3 years the investment is worth $8800. Find the amount invested, correct to the nearest cent.

Step 1 of 2

Write the formula with the unknown amount

8800 = P\left(1 + \dfrac{3}{100}\right)^{3}

Question	Method
a different rate in the first year	$P\left(1 + \dfrac{r_1}{100}\right)\left(1 + \dfrac{r_2}{100}\right)^{n}$
least whole number of years to pass a target	try $n = 1, 2, 3, \ldots$ : the first year past the target

📖 Worked example

Mrs Tan invests $5000 at 2.5% per annum compound interest, compounded yearly. Find the least number of whole years after which the investment is worth more than $6300.

Step 1 of 3

The value after n years

5000 \times 1.025^{n}

Practise this →

5. Standard form

Standard form

\displaystyle A \times 10^{n}, \quad 1 \leq A < 10

$n$ is an integer

Number	Standard form	Point moves
$3\,450\,000$	$3.45 \times 10^{6}$	6 places left
$0.000\,345$	$3.45 \times 10^{-4}$	4 places right
$45\,000$	$4.5 \times 10^{4}$	4 places left
$0.07$	$7 \times 10^{-2}$	2 places right

Multiply: multiply the numbers, add the powers

\displaystyle (a \times 10^{m}) \times (b \times 10^{n}) = ab \times 10^{m + n}

e.g.

\displaystyle (4 \times 10^{3}) \times (5 \times 10^{2})

\displaystyle = 20 \times 10^{5}

\displaystyle = 2 \times 10^{6}

Divide: divide the numbers, subtract the powers

\displaystyle \dfrac{a \times 10^{m}}{b \times 10^{n}} = \dfrac{a}{b} \times 10^{m - n}

e.g.

\displaystyle (9 \times 10^{5}) \div (3 \times 10^{2})

\displaystyle = 3 \times 10^{3}

Add or subtract: same power of 10 first

\displaystyle a \times 10^{n} + b \times 10^{n} = (a + b) \times 10^{n}

e.g.

\displaystyle 3.2 \times 10^{4} + 6.4 \times 10^{3}

\displaystyle = 3.2 \times 10^{4} + 0.64 \times 10^{4}

\displaystyle = 3.84 \times 10^{4}

📖 Worked example

Express $0.000\,345$ in standard form.

Step 1 of 2

Move the decimal point to just after the first non-zero digit

Number of places the point moves:

4

📖 Worked example

Calculate $(3.2 \times 10^{4}) + (6.4 \times 10^{3})$ , giving your answer in standard form.

Step 1 of 2

Write both with the same power of 10

6.4 \times 10^{3} = 0.64 \times 10^{4}

Prefix	Symbol	Power of 10
giga	G	$10^{9}$
mega	M	$10^{6}$
kilo	k	$10^{3}$
milli	m	$10^{-3}$
micro	$\mu$	$10^{-6}$
nano	n	$10^{-9}$

A given power of 10 (not standard form)

\displaystyle 9.6 \times 10^{-13} = 0.96 \times 10^{-12}

Powers with letters

\displaystyle (4 \times 10^{m})(6 \times 10^{n}) = 2.4 \times 10^{m + n + 1}

$24 \times 10^{m + n}$ is not standard form: $24 = 2.4 \times 10^{1}$

📖 Worked example

1 nanometre $(\text{nm}) = 10^{-9}$ metres. Express $450$ $\text{nm}$ in metres, giving your answer in standard form.

Step 1 of 3

Replace the prefix nano by its power of 10

450\ \text{nm} = 450 \times 10^{-9}

metres

Practise this →

Try a question: Indices, compound interest and standard form

Free · no sign-up

Answer a real practice question. Get it wrong and MugCorner shows you where it went wrong, with hints and a step-by-step guide.

Practise indices, compound interest and standard form with step-by-step help

Endless questions from basic to exam level. Get one wrong and see where it went wrong. Free account.

Start free