Logarithm deduction question

Question

Given that $\log_{10}2 = 0.3010$ to four decimal places and that $10^{0.2} < 2$, is it 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.

Can someone walk me through this problem? If you log $10^{0.2}$ with base 10, you end up with $0.2<2$ which is kind of redundant... I've also never been taught about the number of digits thing...

Answer 1

Take common log on $2^{100}$, $$\log 2^{100} = 100 \log 2 \approx 30.10 < 30.2$$ Taking exponent again $$2^{100} < 10^{30.2} = 10^{.2}\times10^{30} < 2\times10^{30}$$

Answer 2

$2^{100}=10^{100*lg2}=10^{30.10}=10^{0.10}\cdot 10^{30}$

We can see that

$2^{100}>1\cdot 10^{30}$ and

$2^{100}<2\cdot 10^{30}$

So $2^{100}$ begins with 1 and has 31 digits.

Answer 3

Saying that a number $N$ has $k$ decimal digits means that the following equalities hold: $$ 10^{k-1}\le N<10^k $$ Taking decimal logarithms, this is equivalent to $$ k-1\le \log_{10}N<k. $$ Since $\log_{10}2\approx0.3010$ (with four exact decimal digits), we can say for sure that $$ \log_{10}2^{100}=100\log_{10}2\approx30.10 $$ with two exact decimal digits. Therefore $2^{100}$ has $31$ decimal digits.

For the initial digit, you need to compute $10^{0.1}\approx1.25$, so the leading digit in $2^{100}$ is $1$. How to compute $10^{0.1}$? Since $1/8>1/10>1/16$, you can do four times the square root of $10$: \begin{align} \sqrt{10}&\approx3.16\\ \sqrt{3.16}&\approx1.77\\ \sqrt{1.77}&\approx1.33\\ \sqrt{1.33}&\approx1.15 \end{align} so you know that $1.15<10^{0.1}<1.33$ and therefore the leading digit of $2^{100}$ is $1$.