Reach for the Stars

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)