Imagine plotting a graph of \(y=2^x\), with \(\quantity{1}{cm}\) to one unit on each axis.
How far along the \(x\)-axis could you go before the graph reached the top of a sheet of paper?
If you extended the graph so the positive \(x\)-axis filled the whole width of a sheet of paper, how tall would the paper have to be?
We could start by drawing up a table of values.
\(x\)
\(y\)
\(0\)
\(1\)
\(1\)
\(2\)
\(2\)
\(4\)
\(3\)
\(8\)
\(4\)
\(16\)
\(5\)
\(32\)
\(6\)
\(64\)
\(7\)
\(128\)
Next we need to decide how big ‘a sheet of paper’ is. If you are in Europe, you are probably working on A4 paper, which is just under \(\quantity{30}{cm}\) on its longest side. From the table of values we can see that the graph extending to \(x=5\) would not quite fit on the page. So the answer is somewhere between \(4\) and \(5\).
For a more accurate answer, we need to find \(x\) such that \(2^x\approx30\).
We could find this by trial and improvement. Alternatively, notice that what we are looking for is \(\log_2{30}\) (the power to which we need to raise \(2\) in order to get \(30\)) which you could evaluate using a calculator. By either method we find that \(x\) is roughly \(4.9\).
If instead we were using American Letter size paper, the sheet would be \(\quantity{11}{in.}\) tall (roughly \(\quantity{279}{mm}\)). How different would the answer be?
A sheet of A2 paper is much bigger - it is twice as tall as A4. How much further along the \(x\)-axis would that allow us to go? Can you explain your answer?
In order to use \(x\) values all the way across a sheet of A4 (\(\quantity{21}{cm}\)), the height of the paper would have to be \(\quantity{2^{21}}{cm} \approx \quantity{20\,971}{m} \approx \quantity{21}{km}\)!
So, \(\quantity{21}{cm}\) across requires the paper to be \(\quantity{21}{km}\) tall. What if the paper were \(\quantity{26}{cm}\) across? Can you explain why this works?
How far along the \(x\)-axis would you have to go so that the graph was tall enough to reach
to the top of The Shard in London?
to the moon?
to the Andromeda galaxy?
Try to estimate the answers before calculating them and mark them at the appropriate points along a sketch of the \(x\)-axis.
Work out where they should be and then add some other results such as the distances to the sun and other stars. What do you notice?
We can use the same techniques as above for these problems, but will have to take care with some unit conversions. We have worked through a couple of examples below. You could choose others from the Data sheet or your own research.
The Shard is roughly \(\quantity{309.6}{m}\) tall, or \(\quantity{30\,960}{cm}\). We could use trial and improvement – from the table of values, our answer must be greater than \(7\) and from our \(\quantity{21}{km}\) piece of paper, it must be less than \(21\). So we could start with a value between \(7\) and \(21\).
Alternatively, using logarithms we need to find \(\log_2{30960}\).
Using either method, we find that the \(x\)-axis would have to extend to roughly \(\quantity{14.9}{cm}\).
The Andromeda galaxy is at a distance of \(\quantity{2.5}{megalight\,years}\) from Earth.
\[\begin{align*}
\quantity{2.5}{Mly}&=\quantity{2.5\times10^6\times365}{light\,days}\\
&=\quantity{2.5\times10^6\times365\times24\times60\times60}{light\,seconds}\\
&=\quantity{2.5\times10^6\times365\times24\times60\times60\times300\,000}{km}\\
&=\quantity{2.5\times10^6\times365\times24\times60\times60\times300\,000\times10^5}{cm}\\
\text{Therefore,}\quad x&=\log_2{(2.5\times10^6\times365\times24\times60\times60\times300\,000\times10^5)}
\end{align*}\]
Now let a calculator deal with the large numbers.
Here is an alternative method, using estimation rather than logarithms.
\[\begin{align*}
\quantity{2.5}{Mly}&=\quantity{2.5\times10^6\times365\times24\times60\times60\times300\,000\times10^5}{cm}\\
&\approx\quantity{2.08\times10^{24}}{cm}\\
&\approx\quantity{2\times1000^8}{cm}\\
&\approx\quantity{2\times(2^{10})^8}{cm}\\
&=\quantity{2^{81}}{cm}\\
\text{Therefore,}\quad x&\approx81
\end{align*}\]
Here is a selection of results arranged along an \(x\)-axis.
Suppose instead of the graph of \(y=2^{x}\) we had used \(y=x^{10}\), or some other power of \(x\). How different would this diagram be?
