Operations with numbers in standard form

Standard form writes any positive number as $a \times 10^{k}$, where $1 \le a < 10$ and $k$ is an integer.

The value of $a$ carries the significant figures; the power of ten carries the size.

So 52 300 000 becomes $5.23 \times 10^{7}$ and 0.000 000 408 becomes $4.08 \times 10^{-7}$.

To find $k$, count how many places the decimal point must move to leave exactly one non-zero digit in front of it.

Moving it left (a big number) gives a positive $k$; moving it right (a small number) gives a negative $k$.

Numbers from 1 up to but not including 10 have $k = 0$, numbers of 10 or more have $k \ge 1$, and numbers below 1 have $k \le -1$.

The condition $1 \le a < 10$ is what makes the form unique.

The number 4800 could be written $48 \times 10^{2}$ or $0.48 \times 10^{4}$, and both are true statements.

Only $4.8 \times 10^{3}$ is standard form.

Note that $a = 10$ is excluded: $10 \times 10^{3}$ is $1 \times 10^{4}$.

Two limiting cases are worth knowing.

Zero cannot be written in this form, because no power of ten makes $a \times 10^{k}$ equal to 0 when $a \ge 1$.

A negative quantity is written with a minus sign in front, such as $-3.2 \times 10^{4}$.

There is a neat link with logarithms, which come later in this topic.

For a number $x \ge 1$, the power $k$ is the whole-number part of $\log_{10} x$.

For example, $\log_{10} 52\,300\,000 = 7.718\ldots$, so $k = 7$.

The order of magnitude of a quantity is this power of ten.

Every significant figure of the number appears in $a$.

To give an answer to 3 significant figures, round $a$ to 3 significant figures and leave $k$ alone.

Watch the one case where rounding changes $k$: $9.996 \times 10^{5}$ to 3 significant figures is $10.0 \times 10^{5}$, which you must rewrite as $1.00 \times 10^{6}$.

The zeros in $1.00$ are significant, so keep them.

Multiply the $a$ values and add the powers; divide the $a$ values and subtract the powers.

This is the exponent law $10^{m} \times 10^{n} = 10^{m+n}$ at work, with $10^{m} \div 10^{n} = 10^{m-n}$.

The result often lands outside $1 \le a < 10$, so finish by adjusting: $24 \times 10^{-4}$ is $2.4 \times 10^{1} \times 10^{-4} = 2.4 \times 10^{-3}$.

The adjusting rule is a balance.

If $a$ is divided by 10, $k$ goes up by 1; if $a$ is multiplied by 10, $k$ goes down by 1.

The value of the number never changes.

A power applies to both factors: $(3 \times 10^{4})^{2} = 3^{2} \times 10^{8} = 9 \times 10^{8}$.

A square root halves the power of ten, so the power must be even first.

To find $\sqrt{1.6 \times 10^{9}}$, rewrite it as $\sqrt{16 \times 10^{8}} = 4 \times 10^{4}$.

Moving one factor of 10 into $a$ is the whole trick, and it is the step most often missed on Paper 1.

You cannot add digits in different place values, so first rewrite one number to share the other's power of ten.

Then add or subtract the $a$ values, and adjust if needed.

For example, $3.6 \times 10^{6} + 4.5 \times 10^{5} = 3.6 \times 10^{6} + 0.45 \times 10^{6} = 4.05 \times 10^{6}$.

When the powers differ by a lot, the smaller number barely changes the larger, which is a useful check.

A GDC shows very large or very small answers as something like 3.901E22.

That is a display convention, not mathematical notation.

A markscheme does not accept it as a final answer, and it does not accept $3.901^{22}$ either.

You must write $3.90 \times 10^{22}$.

On Paper 1 (no calculator) you do every step by hand, and the answer is exact, so $a$ is often a short decimal such as 2.4 or 6.

On Paper 2 the GDC does the arithmetic.

Converting its display into written standard form, and rounding $a$ correctly, is still your job.

Carry full accuracy until the last line; never round an intermediate value to 3 significant figures and reuse it.

Standard form also appears inside other topics.

A sequence or finance question may produce a very large total, and a probability question a very small one.

The same written form and the same accuracy rule apply there.

1. [No GDC] In this question, give each answer in the form $a \times 10^{k}$, where $1 \le a < 10$ and $k \in \mathbb{Z}$.
(a) Write down 63 800 000 in this form. [1]
(b) Write down 0.000 000 405 in this form. [1]
(c) Without a calculator, find $(7 \times 10^{-3}) \times (9 \times 10^{8})$. [2]
(d) Without a calculator, find $\dfrac{2.8 \times 10^{5}}{3.5 \times 10^{-4}}$. [2]

2. [No GDC] Without a calculator, find each value in standard form.
(a) $(6 \times 10^{4})^{3}$ [2]
(b) $\sqrt{4.9 \times 10^{11}}$ [2]

3. [No GDC] Without a calculator, find each value in standard form.
(a) $5.3 \times 10^{-4} - 8 \times 10^{-5}$ [2]
(b) $2.06 \times 10^{9} + 9.4 \times 10^{8}$ [2]

4. [GDC] A grain of a certain pollen has a mass of $2.37 \times 10^{-11}$ kg. A bee carries a load of $1.6 \times 10^{-5}$ kg of this pollen.
(a) Find the number of grains in one load. Give your answer in standard form to 3 significant figures. [2]
(b) Find the mass, in kg, of $3.75 \times 10^{9}$ grains, in standard form to 3 significant figures. [2]

5. [No GDC] Let $x = 4 \times 10^{p}$ and $y = 5 \times 10^{q}$, where $p, q \in \mathbb{Z}$. Find each of the following in the form $a \times 10^{k}$, where $1 \le a < 10$ and $k$ is an expression in $p$ and $q$.
(a) $xy$ [2]
(b) $\dfrac{x}{y}$ [2]
(c) $\dfrac{y^{2}}{x}$ [2]

6. [GDC] A space probe travels $6.25 \times 10^{9}$ km in 9.8 years. Take one year to be 365 days.
(a) Find the probe's mean speed in km per hour, in standard form to 3 significant figures. [3]
(b) Hence find its mean speed in metres per second, in standard form to 3 significant figures. [2]

7. [No GDC] Let $N = (3.2 \times 10^{k}) \times (5 \times 10^{2k})$, where $k \in \mathbb{Z}$.
(a) Show that $N = 1.6 \times 10^{3k+1}$. [2]
(b) Given that $N = 1.6 \times 10^{13}$, write down the value of $k$. [1]
(c) Find the values of $k$ for which $N < 1$. Justify your answer. [2]

8. [No GDC] A student writes the mass of a sample as $25.6 \times 10^{3}$ g.
(a) Explain why this is not in standard form, and write it in standard form. [2]
(b) Arrange $3.1 \times 10^{-4}$, $0.29 \times 10^{-3}$ and $31 \times 10^{-6}$ in increasing order. [2]
(c) Write down how many times larger $3.1 \times 10^{-4}$ is than $31 \times 10^{-6}$. [1]

9. [GDC] Consider $V = \dfrac{8.43 \times 10^{-5}}{(2.9 \times 10^{3})^{2}}$.
(a) Find the value of $V$, giving your answer in standard form to 3 significant figures. [2]
(b) A student writes the answer to (a) as 1.0E-11. State two reasons why this answer does not earn the final mark. [2]

10. [GDC] A library is digitising its collection. At the end of day 1 the archive holds $2.4 \times 10^{12}$ bytes. Each day after that, a further $3.75 \times 10^{11}$ bytes are added, so the amounts at the end of each day form an arithmetic sequence.
(a) Show that at the end of day 5 the archive holds $3.9 \times 10^{12}$ bytes. [2]
(b) Find an expression for the number of bytes at the end of day $n$, in the form $(pn + q) \times 10^{11}$, where $p, q \in \mathbb{R}$. [2]
(c) Hence find the first day on which the archive holds more than $1 \times 10^{14}$ bytes. [3]

On each day, storage costs €$2.5 \times 10^{-12}$ for every byte held at the end of that day.

(d) Find the total storage cost for the first 200 days. Give your answer to the nearest euro. [3]
(e) The library's storage budget is €25 000. Find the last day whose storage can be paid for in full from this budget. [3]