Can we use $\log_{10}$ to find the first digit of $3^{100}$?

Can we use $\log_{10}$ to find the first digit of $3^{100}$?

Question

To nine decimal places, $\log_{10}2=0.301029996$ and $\log_{10}3=0.477121255$.

1. Calculate $\log_{10}5$ and $\log_{10}6$ to three decimal places. By taking logs, or otherwise, show that $5\times 10^{47}<3^{100}<6\times 10^{47}.$

Hence write down the first digit of $3^{100}$.

1. Find the first digit of each of the following numbers: $2^{1000}$; $2^{10\,000}$; and $2^{100\, 000}$.

Solution

The laws of logs tell us that

$$\begin{align*} \log_{10}(ab)&=\log_{10}a+\log_{10}b, \\ \log_{10}(a^b)&=b\log_{10}a, \end{align*}$$

and combining the two we have that $\log_{10}\left(\frac{a}{b}\right)=\log_{10}(ab^{-1})=\log_{10}a+\log_{10}b^{-1}=\log_{10}a-\log_{10}b.$

Recall that $\log_{10}10=1$, by definition. We have that $\log_{10}5=\log_{10}\left(\frac{10}{2}\right)=\log_{10}10-\log_{10}2=1-0.301029996=0.699$ to three decimal places.

Similarly, we have that $\log_{10}6=\log_{10}(3\times 2)=\log_{10}3+\log_{10}2=0.477121255+0.301029996=0.778$ to three decimal places.

Now we wish to show that $5\times 10^{47}<3^{100}<6\times 10^{47}.$ We can calculate that $\log_{10}(5\times 10^{47})=\log_{10}5+47\log_{10}10=0.699+47=47.699.$ Further, we have $\log_{10}3^{100}=100\log_{10}3=47.712,$ to three decimal places. Finally, we find that $\log_{10}(6\times 10^{47})=\log_{10}6+47\log_{10}10=0.778+47=47.778.$ Combining these, we have that $\log_{10}(5\times 10^{47})<\log_{10}3^{100}<\log_{10}(6\times 10^{47}).$ Therefore $10^{\log_{10}(5\times 10^{47})}<10^{\log_{10}3^{100}}<10^{\log_{10}(6\times 10^{47})},$ which is the same as $5\times 10^{47}<3^{100}<6\times 10^{47}.$ From this we can see that $5$ with $47$ zeroes $<3^{100}<$ $6$ with $47$ zeroes. This means that the first digit of $3^{100}$ must be $5$.

To nine decimal places, $\log_{10}2=0.301029996$ and $\log_{10}3=0.477121255$.

1. Find the first digit of each of the following numbers: $2^{1000}$; $2^{10\,000}$; and $2^{100\, 000}$.

To answer this part we mimic the technique of the last part of part (i). That is, we find an integer $n$ which satisfies $n\times10^{m}<2^{1000}<(n+1)\times 10^m$ or $\log_{10}(n\times10^{m})<\log_{10}2^{1000}<\log_{10}((n+1)\times 10^m),$ which tells us that the first digit of $2^{1000}$ is $n$. We have that $\log_{10} 2^{1000}=1000\log_{10}2=301.29996.$ We know that $\log_{10}10^{301}=301,$ so we need to check $\log_{10}(2\times 10^{301})=\log_{10}2+301=301.301.$ Therefore $1\times10^{301}<2^{1000}<2\times10^{301}.$ So the first digit of $2^{1000}$ is $1$.

Again, we study $\log_{10}2^{10\,000}=10\,000\log_{10}2=3010.29996.$ This is again close to $\log_{10}10^{3010}$, so we check $\log_{10}(2\times 10^{3010})=\log_{10}2+3010=3010.301.$ So $10^{3010}<2^{10\,000}<2\times10^{3010},$ which tells us that the first digit of $2^{10\, 000}$ is $1$.

Finally, we have that $\log_{10}2^{100\,000}=100\,000\log_{10}2=30102.9996.$ This is very close to $30103$, so let’s check $\log_{10}(9\times 10^{30102})=\log_{10}3^2+30102=2\log_{10}3+30102=30102.9542.$ So we have that $9\times 10^{30102}<2^{100\,000}<10^{30103},$ and so the first digit of $2^{100\,000}$ is $9$.

This is related to Benford’s Law, which states that the fractional part of the logarithm of random data is uniformly distributed between 0 and 1.

So, for instance, in a set of figures taken from real life, 1 is by far the most common first digit.

This has a neat application in fraud detection, in that most people ignore Benford’s Law when making up figures, and create their first digits in equal numbers.