Title: Default for HELM quiz: Worksheet 1.2
Course: HELM Workbooks 2023-2024

Exercise

Evaluate, without using a calculator,

1. \([value]^3=\)
2. \([value]^5=\)
3. \(2^[value]=\)
4. \([value]^2=\)
5. \([value]^2=\)
Feedback: 1. \([value]^3 = [value]*[value]*[value] = [value]\) 2. \([value]^5 = [value]*[value]*[value]*[value]*[value] = [value]\) 3. To solve \( 2^[value]\) we multiply 2 by itself \([value]\) times to obtain \([value]\) 4. \([value]^2 = [value]*[value] = [value]\) 5. <\([value]^2 = [value]*[value] = [value] \)

Exercise

Evaluate, using a calculator. Give answers to 2 d.p. if necessary.

1. \([value]^[value]=\)
2. \(\displaystyle{[value]^[value]}=\)
Feedback: Both questions can be solved using the \(x^y\) button on your calculator. 1. \( [value]^[value]=[value] \) 2. \(\displaystyle{[value]^[value]}=[value] \)

Recall that when several operations are involved we can make use of the BODMAS rule for deciding the order in which operations must be carried out. The BODMAS rule makes no mention of exponentiation. Exponentiation should be carried out immediately after any brackets have been dealt with and before multiplication and division. Consider the following examples.

Example

Evaluate \(7\times 3^2\).

Solution

There are two operations involved here, exponentiation and multiplication. The exponentiation should be carried out before the multiplication. So \(7\times 3^2 = 7\times 9 = 63 \).

Example

Write out fully

1. \(3m^4\)
2. \((3m)^4\)

Solution

1. In the expression \(3m^4\) the exponentiation is carried out before the multiplication by 3. So \( 3m^4\) means \(3\times(m\times m\times m\times m)\) that is \( 3\times m \times m \times m \times m \)
2. Here the bracketed expression is raised to the power 4 and so should be multiplied by itself four times: \( (3m)^4 = (3m)\times (3m)\times (3m)\times (3m) \)
Because of the associativity of multiplication we can write this as \( 3\times 3\times 3\times 3 \times m\times m \times m \times m \) or simply \(81m^4\).
Note the important distinction between \((3m)^4\) and \(3m^4\).

Index notation

The number \(4\times 4\times 4\) is written, for short, as \(4^3\) and read '4 raised to the power 3' or '4 cubed'.  Note that the number of times '4' occurs in the product is written as a superscript. In this context we call the superscript 3 an index or power. Similarly we could write

- \( 5\times 5  = 5^2\), read '5 to the power 2' or '5 squared', and
- \(7\times 7\times 7\times 7 \times 7 = 7^5,\)
- \(a\times a \times a = a^3, \) and
- \(m\times m \times m \times m = m^4.\)

More generally, in the expression \(x^y\), \(x\) is called the base and \(y\) is called the index or power.  The plural of index is indices. The process of raising to a power is also known as exponentiation because yet another name for a power is an exponent. When dealing with numbers your calculator is able to evaluate expressions involving powers, probably using the \(x^y\) button.

Example

Use a calculator to evaluate \(3^{12}\).

Solution

Using the \(x^y\) button on the calculator check that you obtain \(3^{12}=531441\).

Example

Identify the index and base in the following expressions.

1. \(8^{11}\)
2. \((-2)^5\)
3. \(p^{-q}\)

Solution

1. In the expression \(8^{11}\), 8 is the base and 11 is the index.
2. In the expression \((-2)^{5}\), \(-2\) is the base and 5 is the index.
3. In the expression \(p^{-q}\), \(p\) is the base and \(-q\) is the index.
The interpretation of a negative index will be given later.

Exercise

Write each of the following using index notation:

1. \([value]\times[value]\times[value]\times[value]\times[value]=\)
2. \([value]\times[value]\times[value]\times[value]\times[value]=\)
3. \([value]\times[value]\times[value]\times[value]\times[value]=\)
Feedback: 1. The number \( [value] \) is written 5 times, so \([value]\) is the base and 5 is the index, giving \([value]\) 2. The letter \([value]\) is written 5 times, so \([value]\) is the base and 5 is the index, giving \([value]\) 3. The fraction \([value]\) is written 2 times, so \([value]\) is the base and 2 is the index. Then \([value]\) is written 3 times, so \([value]\) is the base and 3 is the index. Multiplying both expressions together gives \([value]\).

Exercise

Evaluate, without using a calculator. Leave any fractions in fractional form.

1. \(\displaystyle [value]=\)
2. \(\displaystyle [value]=\)
3. \(\displaystyle [value]=\)
4. \(\displaystyle [value]=\)
5. \(\displaystyle [value]=\)
Feedback: 1. \(\displaystyle [value]= [value] \times [value] = [value].\) 2. \(\displaystyle [value]= [value] \times[value] \times[value] =[value].\) 3. \(\displaystyle [value]= [value] \times [value] =[value].\) 4. \(\displaystyle [value]= [value] \times[value] \times [value] =[value].\) 5. \(\displaystyle [value] = [value] \times[value] \times[value] = [value].\)

Laws of indices

There is a set of rules which enable us to manipulate expressions involving indices. These rules are known as the laws of indices, and they occur so commonly that it is worthwhile to memorise them.

Key Point

Laws of Indices
First law | \(a^m \times a^n = a^{m+n}\) | add indices when multiplying numbers with the same base
Second law | \(\dfrac{a^m}{a^n} = a^{m-n}\) | subtract indices when dividing numbers with the same base
Third law | \((a^m)^n = a^{mn}\) | multiply indices together when raising a number to a power

Example

Simplify

1. \(a^5\times a^4\)
2. \(2x^5(x^3)\)

Solution

In each case we are required to multiply expressions involving indices. The bases are the same and we use the first law of indices.

1. The indices must be added, thus \(a^5\times a^4 = a^{5+4}=a^9\).
2. Because of the associativity of multiplication we can write \(2x^5(x^3) = 2(x^5x^3) = 2x^{5+3}=2x^8\).

The first law of indices (above) extends in an obvious way when more terms are involved, as in the following example.

Example

Simplify

1. \(b^5\times b^4\times b^7\)
2. \(y^4y^2y^3\)

Solution

1. The indices are added. Thus \(b^5\times b^4\times b^7=b^{5+4+7}= b^{16}\).
2. All quantities have the same base. To multiply the quantities together, the indices are added: \(y^9\).

Example

Simplify

1. \(\dfrac{8^4}{8^2}\)
2. \(x^{18} \div x^7\)
3. \(\dfrac{5^9}{5^7}\)
4. \(\dfrac{y^5}{y^2}\)
5. \((8^2)^3\)
6. \((z^3)^4\)
7. \((x^2)^5\)
8. \((e^x)^y\)

Solution

In each case we are required to divide expressions involving indices. The bases are the same and we use the second law of indices (above).

1. The indices must be subtracted, thus \(\dfrac{8^4}{8^2} = 8^{4-2}=8^2=64\).
2. Again the indices are subtracted, and so \(x^{18}\div x^7 = x^{18-7} = x^{11}\).
3. The bases are the same, and the division is carried out by subtracting the indices: \(5^{9-7} = 5^2=25\).
4. \(y^{5-2} = y^3\).
5. We use the third law of indices, \((8^2)^3= 8^{2\times 3} = 8^6\).
6. Again, we use the third law of indices, \((z^3)^4= z^{3\times 4} = z^{12}\).
7. \(x^{2\times 5} = x^{10}\).
8. \(e^{x\times y} = e^{xy}\).

Exercise

Write each of the following expressions with a single index:

1. \([value]^[value][value]^[value]=\)
2. \(\dfrac{[value]^[value]}{[value]^[value]}=\)
3. \((x^[value])^[value]=\)
Feedback: 1. We use the first law of indices, so we sum the powers to obtain: \([value]^[value][value]^[value] = [value]^{[value]+[value]} = [value]. \) 2. We use the second law of indices, so we subtract the powers to obtain: \(\dfrac{[value]^[value]}{[value]^[value]} = [value]^{[value]-[value]} = [value]. \) 3. We use the third law of indices, so we multiply both powers to obtain: \((x^[value])^[value]= x^{[value]\times[value]} = [value]. \)

Exercise

Remove the brackets from the following expressions, leaving your answer in the simplest form:

1. \(([value]a)^2=\)
2. \(([value]ab)^3=\)
3. \([value](ab)^3=\)
4. \(([value]xy)^4=\)
Feedback: 1. Removing the brackets we obtain: \( ([value]a)^2 = [value]^2 \cdot a^2 = [value]. \) 2. Removing the brackets we obtain: \( ([value]ab)^3 = [value]^3 \cdot a^3 \cdot b^3 = [value]. \) 3. This time, \([value]\) is not in the brackets, so using the third law of indices we obtain: \( [value](ab)^3 = [value]. \) 4. Again, removing the bracket gives: \( ([value]xy)^4 = [value]^4 \cdot x^4 \cdot y^4 = [value].\)

Two important results which can be derived from the laws of indices state:

Key Point

- Any non-zero number raised to the power \(0\) has the value \(1\) , that is \(a^0=1\).
- Any number raised to power \(1\) is itself, that is \(a^1=a\).

A generalisation of the third law of indices states:

Key Point

\((a^mb^n)^k = a^{mk}b^{nk}\) .

Example

Remove the brackets from

1. \((3x)^2\)
2. \((x^3y^7)^4\)

Solution

1. Noting that \(3=3^1\) and \(x=x^1\) then \((3x)^2= (3^1x^1)^2 = 3^2x^2=9x^2\).
Alternatively \((3x)^2=(3x)\times (3x) = 9x^2\).
2. \((x^3y^7)^4 = x^{3\times 4}y^{7\times 4} = x^{12}y^{28}\).

Exercise

Simplify, leaving your answer in the simplest form:

1. \([value]x^[value](x^[value])=\)
2. \([value]x^[value]([value]x)=\)
3. \([value]x^{-1}([value]x^[value])= \)
4. \([value]x(x^[value])=\)
Feedback: 1. Removing the brackets and using the first law of indices we get: \( [value]x^[value](x^[value]) = [value] \cdot x^{[value]+[value]} = [value].\) 2. Multiplying like terms and using the first law of indices we get: \( [value]x^[value]([value]x) = [value] \cdot [value] \cdot x^[value] \cdot x = [value] \cdot [value] \cdot x^{[value]+1} = [value] .\) 3. Similarly, \([value]x^{-1}([value]x^[value])= [value] \cdot [value] \cdot x^{-1} \cdot x^[value] = [value] \cdot [value] \cdot x^{[value] - 1} = [value]. \) 4. Again, we obtain: \( [value]x(x^[value])= [value] \cdot x^{1+[value]} = [value].\)

Exercise

Simplify, leaving your answer in the simplest form:

1. \([value]x^[value](x^[value])=\)
2. \([value]x^[value](x^[value])=\)
3. \([value]x^{[value]}(x^[value])= \)
4. \([value]x^[value]([value]x^[value])=\)
Feedback: 1. Removing the brackets and using the first law of indices, we obtain: \( [value]x^[value](x^[value]) = [value] \cdot x^{[value]+[value]} = [value]. \) 2. Similarly, \( [value]x^[value](x^[value]) = [value] \cdot x^{[value]+[value]} = [value]. \) 3. Again, \( [value]x^{[value]}(x^[value])= [value] \cdot x^{[value]+[value]} = [value]. \) 4. Finally multiplying like terms and using the first law of indices, we get: \( [value]x^[value]([value]x^[value]) = [value] \cdot [value] \cdot x^{[value]+[value]} = [value].\)

Polynomial expressions

An important group of mathematical expressions which use indices are known as polynomials.

Examples of polynomials are

- \(4x^3+2x^2+3x-7\),
- \( x^2+x \),
- \( 17-2t+7t^4 \),
- \( z-z^3 \).

Notice that they are all constructed using non-negative whole number powers of the variable. Recall that \(x^0=1\) and so the number \(-7\) appearing in the first expression can be thought of as \(-7x^0\). Similarly the \(17\) appearing in the third expression can be read as \(17t^0\).

Key Point

A polynomial expression takes the form

\( a_0 + a_1x + a_2x^2 + a_3x^3 +\ldots + a_nx^n\)

where \( a_0 ,a_1, a_2, a_3, \ldots a_n\) are all constants called the coefficients of the polynomial. The number \(a_0\) is also called the constant term. The highest power in a polynomial is called the degree of the polynomial.

Polynomials with low degrees have special names and subscript notation is often not needed:

Polynomial | Degree | Name
\[ ax^3+bx^2+cx+d \] | 3 | cubic
\[ ax^2+bx+c \] | 2 | quadratic
\[ ax+b \] | 1 | linear
\[ a\] | 0 | constant

Exercise

Which of the following expressions are polynomials? Give the degree of those which are, put \(0\) if they are not.

 | polynomial | degree
\( 3x^2+4x+2 \) |  |
\( \dfrac{1}{x+1} \) |  |
\( \sqrt{x} \) |  |
\( 2t+4 \) |  |
\( 3x^2+\dfrac{4}{x}+2 \) |  |

Recall that a polynomial expression must contain only terms involving non-negative whole number powers of the variable.
Feedback: Recall a polynomial takes the form: \( a_0 + a_1.x + a_2.x^2 + ... \) and the \(\textbf{degree}\) of a polynomial is the \( \textbf{highest power} \) in the polynomial.

Exercise

State which of the following are linear polynomials, which are quadratic polynomials, and which are constants. (Recall that a polynomial expression must contain only terms involving non-negative whole number powers of the variable.)

\( x \) |
\( x^2+x+3 \) |
\( x^2-1 \) |
\( 3-x \) |
\( 7x-2 \) |
\( \dfrac{1}{2} \) |
\( \dfrac{1}{2}x+\dfrac{3}{4} \) |
\( 3-\dfrac{1}{2}x^2 \) |
Feedback: Recall what each of the terms mean: \( \textbf{Linear:} \) The highest power in the polynomial is \(1\). \( \textbf{Quadratic:} \) The highest power in the polynomial is \(2\). \( \textbf{Constant:} \) There is only one constant term \(a_0\).

Exercise

State which of the following are polynomials.

\( -\alpha^2-\alpha-1 \) |
\( x^{1/2}-7x^2 \) |
\( \dfrac{1}{x} \) |
\( 19 \) |
\( 4t+17 \) |
\( \dfrac{1}{2}-\dfrac{1}{2}t \) |
\( 15 \) |
\( t^2-3t+7 \) |
\( \dfrac{1}{t^2}+\dfrac{1}{t}+7 \) |
Feedback: Remember a polynomial expression takes the form \( a_0 + a_1.x + a_2.x^2 + a_3.x^3 + ... + a_n.x^n \).

Negative indices

Sometimes a number is raised to a negative power. This is interpreted as follows:

Key Point

Negative Powers

\(a^{-m} = \dfrac{1}{a^m}, \qquad a^m = \dfrac{1}{a^{-m}}\)

Thus a negative index can be used to indicate a reciprocal.

Example

Write each of the following expressions using a positive index and simplify if possible.

1. \(2^{-3}\)
2. \(\dfrac{1}{4^{-3}}\)
3. \(x^{-1}\)
4. \(x^{-2}\)
5. \(10^{-1}\)

Solution

1. \(2^{-3} = \dfrac{1}{2^3} = \dfrac{1}{8}\)
2. \(\dfrac{1}{4^{-3}} = 4^3 = 64\)
3. \(x^{-1}= \dfrac{1}{x^1}=\dfrac{1}{x}\)
4. \(x^{-2}= \dfrac{1}{x^2}\)
5. \(10^{-1}=\dfrac{1}{10^1}= \dfrac{1}{10} \) or \( 0.1\)

Exercise

Write the following numbers using a positive index and also express your answers as decimal fractions:

1. As a fraction, \([value]= \)
As a decimal, \([value]= \)

2. As a fraction, \([value]= \)
As a decimal, \([value]= \)

3. As a fraction, \([value]= \)
As a decimal, \([value]= \)
Feedback: 1. Using the fact that \( a^{-m} = \dfrac{1}{a^m} \) we can see \([value] = \displaystyle [value]. \) Now, converting the answer from a fraction to a decimal we obtain \(\displaystyle[value] = [value].\) 2. Using the fact that \( a^{-m} = \dfrac{1}{a^m} \) we can see \([value] = \displaystyle[value] .\) Now, converting the answer from a fraction to a decimal we obtain \(\displaystyle[value] = [value].\) 3. Using the fact that \( a^{-m} = \dfrac{1}{a^m} \) we can see \([value] = \displaystyle[value] .\) Now, converting the answer from a fraction to a decimal we obtain \(\displaystyle[value] = [value].\)

Exercise

Use the laws of indices to simplify.

1. \(\dfrac{a^[value] \times a^[value]}{a^[value]}= \)
2. \(\dfrac{m^[value] \times m^[value]}{m^[value]}=\)
Feedback: 1. Use the first law of indices to simplify the numerator: \(\dfrac{a^[value] \times a^[value]}{a^[value]}= \dfrac{a^{[value]+[value]}}{a^[value]}= \dfrac{a^{[value]}}{a^[value]}.\) Now use the second law to simplify the result: \(\dfrac{a^{[value]}}{a^[value]}= a^{[value]-[value]} = a^[value].\) 2. Use the first law of indices to simplify the numerator: \(\dfrac{m^[value] \times m^[value]}{m^[value]}= \dfrac{m^{[value]+[value]}}{m^[value]}= \dfrac{m^{[value]}}{m^[value]}.\) Now use the second law to simplify the result: \(\dfrac{m^{[value]}}{m^[value]}= m^{[value]-[value]} = m^[value].\)

Exercise

Use the laws of indices to simplify.

1. \(x^[value] \times x^[value]= \)
2. \(\dfrac{t^[value]}{t^[value]}=\)
3. \(\dfrac{y^[value]}{y^[value]}=\)
Feedback: 1. Use the first law of indices: \(x^[value] \times x^[value]= x^{[value]+[value]} =x^[value]= x.\) 2. Use the second law of indices: \(\dfrac{t^{[value]}}{t^[value]}= t^{[value]-([value])} = t^[value].\) 3. Use the second law of indices, as for (b). \(\dfrac{y^{[value]}}{y^[value]}= y^{[value]-([value])} = y^[value].\)

Fractional indices

So far we have used indices that are whole numbers. We now consider fractional powers. Consider the expression \((16^{1\over 2})^2\). Using the third law of indices, \((a^m)^n=a^{mn}\), we can write

\( (16^{1\over 2})^2 = 16^{{1\over 2}\times 2} = 16^1 =16\).

So \(16^{1\over 2}\) is a number which when squared equals 16, that is 4 or \(-4\). In other words \(16^{1\over 2}\) is a square root of 16. There are always two square roots of a non-zero positive number, and we write \(\ 16^{1\over 2} = \pm 4\).

Key Point

In general, \(a^{\frac{1}{2}}\) is a square root of \(a\) where \(a>0\)

Similarly

\( (8^{1\over 3})^3 = 8^{{1\over 3}\times 3} = 8^1 = 8\)

so that \(8^{1\over 3}\) is a number which when cubed equals \(8\). Thus \(8^{1\over 3}\) is the cube root of \(8\), that is \(\sqrt[3]{8}\), namely \(2\). Each number has only one cube root, and so

\( 8^{1\over 3} = 2\).

Key Point

In general, \(a^{\frac{1}{3}}\) is the cube root of \(a\)

More generally we have

Key Point

The \(n\)th root of \(a\) is denoted by \(a^{\frac{1}{n}}\).

When \(a<0\) the \(n\)th root only exists if \(n\) is odd.

If \(a>0\) the positive \(n\)th root is denoted by \(\sqrt[n]{a}\)

If \(a<0\) the negative \(n\)th root is \(-\sqrt[n]{|a|}\)

Your calculator will be able to evaluate fractional powers, and roots of numbers. Check that you can obtain the results of the following Examples on your calculator, but be aware that calculators normally give only one root when there may be others.

Example

Evaluate.

1. \(144^{1/2}\)
2. \(125^{1/3}\)

Solution

1. \(144^{1/2}\) is a square root of 144, that is \(\pm 12.\)
2. Noting that \(5^3=125\), we see that \(125^{1/3}=\sqrt[3]{125}=5.\)

Example

Evaluate.

1. \(32^{1/5}\)
2. \(32^{2/5}\)
3. \(8^{2/3}\)

Solution

1. \(32^{1\over 5}\) is the 5th root of 32, that is \(\sqrt[5]{32}\). Now \(2^5=32\) and so \(\sqrt[5]{32}=2\).
2. Using the third law of indices we can write \(32^{2/5}=32^{2\times {1\over 5}} = (32^{1\over 5})^2\). Thus \( 32^{2/5}=((32)^{1/5})^2=2^2=4\).
3. Note that \(8^{1/3}=2\). Then \( 8^{2\over 3} = 8^{2\times {1\over 3}} = (8^{1/3})^2 = 2^2=4\).
Note the following alternatives: \( 8^{2/3} = (8^{1/3})^2 = (8^2)^{1/3}\).

Exercise

Evaluate, using a calculator, giving positive answers only. Where your answer is a decimal number, give it to \(4\) significant figures.

1. \([value]^{\frac{1}{2}}=\)
2. \([value]^{-\frac{1}{3}}=\)
3. \([value]^3=\)
4. \([value]^{\frac{1}{4}}=\)
Feedback: 1. \([value]^{\frac{1}{2}}= [value]\) 2. \([value]^{-\frac{1}{3}}= [value]\) 3. \([value]^3= [value]\) 4. \([value]^{\frac{1}{4}}= [value]\)

Exercise

Evaluate, using a calculator, giving positive answers only. Give your answers to \(4\) significant figures.

1. \([value]^{-5}=\)
2. \([value]^{-\frac{2}{7}}=\)
Feedback: 1. \([value]^{-5}= [value]\) 2. \([value]^{-\frac{2}{7}}= [value]\)

Example

Write the following as a simple power with a single index.

1. \(\sqrt{x^5}\)
2. \(\sqrt[4]{x^3}\)

Solution

1. \(\sqrt{x^5} = (x^5)^{\frac{1}{2}}\). Then using the third law of indices we can write this as \(x^{5\times{\frac{1}{2}}}=x^{\frac{5}{2}}.\)
2. \(\sqrt[4]{x^3} =(x^3)^{\frac{1}{4}}\). Using the third law we can write this as \(x^{3\times \frac{1}{4}}=x^{\frac{3}{4}}.\)

Example

Show that \(z^{-\frac{1}{2}}=\dfrac{1}{\sqrt{z}}.\)

Solution

\(z^{-\frac{1}{2}}=\dfrac{1}{z^{\frac{1}{2}}}=\dfrac{1}{\sqrt{z}}.\)

Example

The generalisation of the third law of indices states that \((a^mb^n)^k=a^{mk}b^{nk}\). By taking \(m=1\), \(n=1\) and \(k=\dfrac{1}{2}\) show that \(\sqrt{ab}=\sqrt{a}\,\sqrt{b}\).

Solution

Taking \(m=1\), \(n=1\) and \(k=\dfrac{1}{2}\) gives

\((ab)^{\frac{1}{2}}=a^{\frac{1}{2}}b^{\frac{1}{2}}\).

Taking the case when all these roots are positive, we have

\(\sqrt{ab}=\sqrt{a}\sqrt{b}\)

Key Point

\(\sqrt{ab}=\sqrt{a}\sqrt{b} \qquad  a\ge 0,\ \ b\ge 0\).

This result often allows answers to be written in alternative forms. For example, we may write \(\sqrt{48}\) as \(\sqrt{3\times 16}=\sqrt{3}\sqrt{16}=4\sqrt{3}\).

Although this rule works for multiplication we should be aware that it does not work for addition or subtraction so that \(\sqrt{a\pm b}\ne \sqrt{a}\pm\sqrt{b}\).

Exercise

Simplify, writing your answers using a single index.

1. \(\dfrac{a^[value]a^{\frac{3}{4}}}{a^{-\frac{1}{2}}}= \)
2. \(\dfrac{\sqrt{z}}{z^{\frac[value]{2}}}= \)
3. \(\dfrac{z^{-\frac[value]{2}}}{\sqrt{z}}= \)
4. \(\dfrac{\sqrt[[value]]{a}}{\sqrt[[value]]{a}}= \)
5. \(\dfrac{\sqrt[[value]]{z}}{z^{\frac{1}{2}}}= \)
Feedback: FEEDBACK 1. Use the first and second laws of indices: \(\dfrac{a^[value]a^{\frac{3}{4}}}{a^{-\frac{1}{2}}}= a^{[value]+\frac{3}{4}+\frac{1}{2}}=[value]\). 2. Change the \(\sqrt{z}\) to index notation: \(\dfrac{\sqrt{z}}{z^{\frac[value]{2}}} = \dfrac{z^{\frac{1}{2}}}{z^{\frac[value]{2}}}\). Now use the second law to simplify the result: \(\dfrac{z^{\frac{1}{2}}}{z^{\frac[value]{2}}}=z^{\frac{1}{2}-\frac[value]{2}}=[value]\). 3. Change the \(\sqrt{z}\) to index notation and then use the second law of indices: \(\dfrac{z^{-\frac[value]{2}}}{\sqrt{z}}= \dfrac{z^{-\frac[value]{2}}}{z^{\frac{1}{2}}}=z^{-\frac[value]{2}-\frac[value]{2}}=[value]\). 4. As above: \(\dfrac{\sqrt[[value]]{a}}{\sqrt[[value]]{a}}=a^{\frac{1}[value] - \frac{1}[value]}=[value]\). 5. And again: \(\dfrac{\sqrt[[value]]{z}}{z^\frac{1}{2}}=z^{\frac{1}[value] -\frac{1}{2}}=[value]\).

Exercise

Write each of the following expressions with a single index:

1. \( (x^[value])^[value]= \)
2. \(x^[value]x^[value]= \)
3. \(\dfrac{x^[value]}{x^[value]}= \)
Feedback: 1. Use the third law of indices: \((x^[value])^[value]= x^{[value]\times[value]}=[value]\). 2. Use the first law of indices: \(x^[value]x^[value] = x^{[value]+[value]}=[value]\). 3. Use the second law of indices: \(\dfrac{x^[value]}{x^[value]}=x^{[value]-[value]}=[value]\).

Scientific notation

It is often necessary to use very large or very small numbers such as \(78000000\) and \(0.00000034\).

Scientific notation can be used to express such numbers in a more concise form. Each number is written in the form

\( a\times 10^n\) where \(a\) is a number between \(1\) and \(10\).

We can make use of the following facts:

- \(10=10^1, \quad 100=10^2, \quad 1000=10^3\) and so on,
- \( 0.1=10^{-1},\quad 0.01=10^{-2},\quad 0.001=10^{-3}\) and so on.

For example,

- the number \(5000\) can be written \(5\times 1000=5\times 10^3\),
- the number \(403\) can be written \(4.03\times 100=4.03\times 10^2\),
- the number \(0.009\) can be written \(9\times 0.001=9\times 10^{-3}\).

Furthermore, to multiply a number by \(10^n\) the decimal point is moved \(n\) places to the right if \(n\) is a positive integer, and \(n\) places to the left if \(n\) is a negative integer. (If necessary additional zeros are inserted to make up the required number of digits before the decimal point.)

Example

Write the following numbers in scientific notation.

1. \(0.00678\)
2. \(123456.7\)

Solution

1. \(0.00678=6.78\times 10^{-3}\)
2. \(123456.7=1.234567\times 10^5\)

Engineering constants

Many constants appearing in engineering calculations are expressed in scientific notation. For example the charge on an electron equals \(1.6\times 10^{-19}\) coulomb and the speed of light is \(3\times 10^8\) m s\(^{-1}\). Avogadro’s constant is equal to \(6.023\times 10^{26}\) and is the number of atoms in one kilomole of an element. Clearly the use of scientific notation avoids writing lengthy strings of zeros.

Your scientific calculator will be able to accept numbers in scientific notation. Often the \(E\) button is used and a number like \(4.2\times 10^7\) will be entered as \(4.2 E7\). Note that \(10E4\) means \(10\times 10^4\), that is \(10^5\). To enter the number \(10^3\) say, you would key in \(1E3\). Entering powers of 10 incorrectly is a common cause of error. You must check how your particular calculator accepts numbers in scientific notation.

The following example is designed to check that you can enter numbers given in scientific notation into your calculator.

Example

Use your calculator to find \(4.2\times 10^{-3}\times 3.6\times 10^{-4}\).

Solution

\(1.512 \times 10^{−6}\).

Exercise

Express each of the following numbers in scientific notation. Please give your answer in the form \(a\times 10^m\), and not in the form \(aEm\).

1. \([value]=\)
2. \([value]=\)
3. \([value]=\)
4. \([value]=\)
5. \([value]=\)
6. \([value]=\)
7. \([value]=\)
Feedback: 1. \([value]\) can be written as \( [value] \times 10 = [value].\) 2. \([value]\) can be written as \( [value] \times 100 = [value].\) 3. \([value]\) can be written as \( [value] \times 1000 = [value].\) 4. \([value]\) can be written as \( [value] \times 1000000 = [value].\) 5. \([value]\) can be written as \( 2 \times 0.1 = [value].\) 6. \([value]\) can be written as \( 3.4 \times 0.01 = [value].\) 7. \([value]\) can be written as \( 9.856 \times 0.01 = [value].\)

Exercise

Simplify \([value]\times 10^[value]\times[value]\times 10^[value]\).
Feedback: We multiply the two numbers \([value]\) and \([value]\) then using our first law of indices we add the powers of 10. So we get \([value]\times [value]\times 10^{[value]+([value])} = [value]. \)