**Standard Index Form**

Standard index form is also known as standard form. It is very useful when writing very big or very small numbers.

In standard form, a number is always written as: a × 10 ^{n}

a is always between 1 and 10. n tells us how many places to move the decimal point.

**Example :**** **Write 15 000 000 in standard index form.

**Solution**

15 000 000 = 1.5 × 10 000 000

This can be rewritten as:

1.5 × 10 × 10 × 10 × 10 × 10 × 10 × 10

= 1. 5 × 10 ^{7}

You can convert from standard form to ordinary numbers, and back again. Have a look at these examples:

3 x 10^{4} = 3 × 10 000 = 30 000 (Since 10^{4} = 10 × 10 × 10 × 10 = 10 000)

2 850 000 = 2.85 × 1 000 000 = 2.85 × 10^{6} Make the first number between 1 and 10.

0.000467 = 4.67 × 0.0001 = 4.67 × 10 ^{-4}

**Adding and subtracting numbers in standard index form**

Convert them into ordinary numbers, do the calculation, then change them back if you want the answer in standard form.

**Example 1**

4.5 × 10^{4} + 6.45 × 10^{5}

= 45,000 + 645,000

= 690,000

= 6.9 × 10^{5}

**Multiplying and dividing numbers in standard form:**

Here you can use the rules for multiplying and dividing powers. Remember these rules:

To multiply powers you add, eg, 10^{5} × 10^{3} = 10^{8}

To divide powers you subtract, eg, 10^{5} ÷ 10^{3} = 10^{2}

**Example 2**

Simplify (2 × 10^{3}) × (3 × 10^{6})

**Solution**

Multiply 2 by 3 and add the powers of 10:

(2 × 10^{3}) × (3 × 10^{6}) = 6 × 10^{9}

Question

Simplify (36 × 10^{5}) ÷ (6 × 10^{3})

**Answer**

Did you get 6 ×10^{2}?

If not, remember that you should first work out 36 ÷ 6, then work subtracts the powers of 10 (because it is division), like this:

(36 x 10^{5}) ÷ (6 × 10^{3}) = (36 ÷ 6) x (10^{5} ÷ 10^{3}) = 6 ×10^{2}

**Mixed Problem and standard form**

In science and astronomy, many measurements use very small and very large numbers. For this reason, most scientists prefer to do calculations in standard form. Scientists use standard form so often that its sometimes called **scientific notation.**

Example

The density of hydrogen is 8.89 X 10^{-5} g/cm^{3 }

- Find the mass of 1 m
^{3 }of hydrogen. - Argon is approximately 20 times as dense as hydrogen. Find the density of argon, giving the answer in standard form correct to 3 s.f.

Solution

- 1 m
^{3 }= 10^{6 }cm^{3 }

mass of 1 m^{3 }hydrogen = 8.89 X 10^{-5 }X 10^{6} g

= 8.89 X 10^{1 }g

= 88.9 g

- density of argon

= 20 X 8.89 X 10^{-5} g/cm^{3}

= 2 X 10^{1 }X 8.89 X 10^{-5} g/cm^{3}

= 2 X 8.89 X 10-4 g/cm^{3 }

= 17.78 X 10^{-4 }g/cm^{3}

= 1.778 X 10^{-3 }g/cm^{3 }

= 1.78 X 10^{-3 }g/cm to 3 s.f.