Review question

# How large is $2^{100}$, and what is its starting digit? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R6045

## Solution

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;

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

3. $2^{100}$ begins in a $1$ and is $31$ digits long;

4. $2^{100}$ begins in a $2$ and is $31$ digits long.

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

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