How large is $2^{100}$, and what is its starting digit?

How large is $2^{100}$, and what is its starting digit?

Question

Given that $\log_{10} 2 = 0.3010 \; \text{ (to }4\text{ d.p.)}$ and that $10^{0.2} < 2$ it is possible to deduce that

1. $2^{100}$ begins in a $1$ and is $30$ digits long;

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1. $2^{100}$ begins in a $2$ and is $30$ digits long;

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1. $2^{100}$ begins in a $1$ and is $31$ digits long;

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1. $2^{100}$ begins in a $2$ and is $31$ digits long.

Solution

From the first given property, we have that $\log_{10} 2^{100} = 100\log_{10} 2 = 30.10\dots c$ so $2^{100} = 10^{30.10 \dots c} = 10^{0.10\dots c} \times 10^{30}$ and so $2^{100}$ is $31$ digits long, with the first digit being the first digit of $10^{0.10\dots c}$.

Now we use the second given property: $1 < 10^{0.10\dots c} < 10^{0.2} < 2.$ The first digit is less than $2$ and so must be a $1$, and the answer is (c).