Reach for the stars
Some graphs grow so quickly you can reach dizzy heights...
Age
16 to 18
Challenge level
Problem
Imagine plotting a graph of $y=2^x$, with $1$cm to one unit on each axis.
Image
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?
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 have provided some data below for you to work from, or you could research suitable data for yourself.
Paper sizes
* A4 paper measures $298$mm by $210$mm.
* A3 is double the area of A4 --- $298$mm by $420$mm.
* To go from A4 to A3 and from A3 to A2 you double the shorter dimension each time. The sequence continues up to A0.
* American Letter size paper is $8.5$in. by $11$in..
On Earth
* The Shard in London is $309.6$m high. Its viewing gallery is approximately $244$m above ground level.
* One of the tallest buildings on Earth is the Burj Khalifa in Dubai, at $829.8$m.
* The highest mountain is Mount Everest, at $8\,848$m.
In astronomy
* The mean distance from the Earth to the moon is $378\,000$km.
* The mean distance from the Earth to the sun is $149\,600\,000$km.
* After the Sun, the nearest bright star is Alpha Centauri, at a distance of about $4.4$light years from Earth.
One light year is the distance that light can travel in a year. Light travels at a speed of about $300\,000$km/s.
* Andromeda, the closest major galaxy to our own, is approximately $2.5$Mly from Earth.
$1$Mly (megalight year) is $10^6$light years.
* The edge of the known Universe is approximately $13.8$ billion light years away.
* A4 paper measures $298$mm by $210$mm.
* A3 is double the area of A4 --- $298$mm by $420$mm.
* To go from A4 to A3 and from A3 to A2 you double the shorter dimension each time. The sequence continues up to A0.
* American Letter size paper is $8.5$in. by $11$in..
On Earth
* The Shard in London is $309.6$m high. Its viewing gallery is approximately $244$m above ground level.
* One of the tallest buildings on Earth is the Burj Khalifa in Dubai, at $829.8$m.
* The highest mountain is Mount Everest, at $8\,848$m.
In astronomy
* The mean distance from the Earth to the moon is $378\,000$km.
* The mean distance from the Earth to the sun is $149\,600\,000$km.
* After the Sun, the nearest bright star is Alpha Centauri, at a distance of about $4.4$light years from Earth.
One light year is the distance that light can travel in a year. Light travels at a speed of about $300\,000$km/s.
* Andromeda, the closest major galaxy to our own, is approximately $2.5$Mly from Earth.
$1$Mly (megalight year) is $10^6$light years.
* The edge of the known Universe is approximately $13.8$ billion light years away.
This is an Underground Mathematics resource.
Underground Mathematics is hosted by Cambridge Mathematics. The project was originally funded by a grant from the UK Department for Education to provide free web-based resources that support the teaching and learning of post-16 mathematics.
Visit the site at undergroundmathematics.org to find more resources, which also offer suggestions, solutions and teacher notes to help with their use in the classroom.
Underground Mathematics is hosted by Cambridge Mathematics. The project was originally funded by a grant from the UK Department for Education to provide free web-based resources that support the teaching and learning of post-16 mathematics.
Visit the site at undergroundmathematics.org to find more resources, which also offer suggestions, solutions and teacher notes to help with their use in the classroom.
Student Solutions
Pablo from King's College Alicante has found approximate and accurate distances along the $x$-axis for when the height is the height of A4 paper, the height of the Shard, the distance to the moon, and the distance to the edge of the observable universe. Here are his solutions.
A4 Paper
An A4 piece of paper is about $29.8 \, \text{cm}$ by $21 \, \text{cm}$. If the scale of the graph is 1 unit : 1 cm, then the $x$-coordinate satisfies
$2^x=29.8$
To estimate this, you can see that
$16< 29.8 < 32$
$2^4< 29.8 < 2^5$.
$2^4<2^x<2^5$
Therefore $4 < x < 5$
To find the actual value, we can calculate $\log_2{29.8}=4.90 \, \text{cm}$ (2dp)
If the positive $x$-axis was equal to the width then
$y=2^{21}$ (as the width of an A4 is $21 \, \text{cm} $)
$2^{10}$ is approximately equal to $1,000$, so $2^{20}$ is approximately equal to $1,000,000$. Therefore $2^{21}$ is approximately equal to $2,000,000$ ($1,000,000 \times 2$). The actual value is $2,097,152 \, \text{cm} = 20.97152 \, \text{km}$
The Shard
The height of The Shard is $309.6 \, \text{m} = 30960 \, \text{cm}$
$30960$ is approximately $32 \times 1000 = 2^{5} \times 2^{10} = 2^{15}$. So the $x$-axis would have to be $15 \, \text{cm} $ long.
The exact value is $14.91 \text{cm}$.
The moon
The average distance from the earth to the moon is $378,000 \, \text{km}$. This is equal
to $37,800,000,000 \, \text{cm} $
This means that $2^x=3.78 \times 10^{10}$
This is about $32 \, \text{billion} = 2^{5} \times 2^{10} \times 2^{10} \times 2^{10}$ (as $2^{10}$ is
approximately equal to $1,000$)
Therefore $x$ is about $5+10+10+10=35 \, \text{cm}$
The actual value is $\log_2{3.78 \times 10^{10}} = 35.14 \, \text{cm}$(2dp)
The edge of observable universe
Finally the edge of the known universe is $13.8 \, \text{bly}$ away.
Light travels at $300,000 \, \text{km/s}$
To turn this into $\text{km/yr}$ (i.e.ly), we multiply $300,000 \times 60 \times 60 \times 24 \times 365$
(Seconds->Minutes->Hours->Days->Years)
This is approximately equal to $9.5 \times 10^{12} \, \text{km}$.
$9.5 \times 10^{12} \, \text{km} = 9.5 \times 10^{17} \, \text{cm}$
$13.8 \, \text{billion}$ is roughly equal to $14 \times 10^{9}$
$(14 \times 10^9) \times (9.5 \times 10^{17}) = 133 \times 10^{26} = 1.33 \times 10^{28} \text{cm}$
$1.33 \times 10^{28}$ is roughly equal to $10^{27}=(10^3)^9$. Since $2^{10}$ is roughly
$10^3$, then $1.33 \times 10^{28}$ is approximately equal to $(2^{10})^9$
So the $x$-axis should be about $9 \times 10=90 \, \text{cm} $ long
The actual value is $\log_2{1.33 \times 10^{28}} = 93.42 \, \text{cm} $ (2dp)