How do you solve log_10 2?

Mar 13, 2016

${\log}_{10} 2 \approx 0.30103$

Explanation:

One somewhat cumbersome way of calculating ${\log}_{10} 2$ goes as follows:

First note that $2 < {10}^{1}$, so ${\log}_{10} 2 < 1$ and we can write $\textcolor{b l u e}{0.}$ as the start of the logarithm.

If we raise $2$ to the $10$th power, then the effect on the logarithm is to multiply it by $10$.

Let's see what happens:

${2}^{10} = 1024 \ge 1000 = {10}^{3}$

So the first digit after the decimal point is $\textcolor{b l u e}{3}$

Divide $\frac{1024}{10} ^ 3$ to get $1.024$

To find the next digit of the logarithm, calculate ${1.024}^{10}$ and compare it with powers of $10$:

${1.024}^{10} \approx 1.26765060022822940149$

This is still less than ${10}^{1}$, so the next digit of the logarithm is $\textcolor{b l u e}{0}$

Then:

${1.26765060022822940149}^{10} \approx 10.71508607186267320891 \ge {10}^{1}$

So the next digit of the logarithm is $\textcolor{b l u e}{1}$

Divide $10.71508607186267320891$ by ${10}^{1}$ to get:
$1.071508607186267320891$

Then:

${1.071508607186267320891}^{10} \approx 1.99506311688075838379$

This is still less than ${10}^{1}$, so the next digit of the logarithm is $\textcolor{b l u e}{0}$

Then:

${1.99506311688075838379}^{10} \approx 999.00209301438450246726$

That is very close to $1000 = {10}^{3}$, so a very good approximation is to stop here with a digit $\textcolor{b l u e}{3}$, bearing in mind that it would actually be very slightly less.

Putting our digits together ${\log}_{10} 2 \approx 0.30103$