# How do you evaluate log 2?

Oct 24, 2015

One (somewhat impractical) way is to raise to the $10$th power repeatedly and compare with powers of $10$ as follows:

#### Explanation:

$2 < {10}^{1}$ so $\log \left(2\right) < 1$ and the portion of $\log \left(2\right)$ before the decimal point is: $\textcolor{red}{0}$

If we raise $2$ to the $10$th power, then the $\log$ of the resulting number will be $10 \log \left(2\right)$, so

${10}^{3} = 1000 < {2}^{10} = 1024 < {10}^{4} = 10000$

So the first digit of $\log \left(2\right)$ after the decimal point is: $\textcolor{red}{3}$

$\frac{1024}{{10}^{3}} = 1.024$

To find the next decimal place, evaluate ${1.024}^{10}$ and compare it to powers of $10$ to find:

${10}^{0} < {1.024}^{10} \approx 1.2676506 < {10}^{1}$

So the next decimal place is: $\textcolor{red}{0}$

Then:

${10}^{1} < {1.2676506}^{10} \approx 10.71508605 < {10}^{2}$

So the next decimal place is: $\textcolor{red}{1}$

Divide $10.71508605$ by ${10}^{1}$ then find:

${10}^{0} < {1.071508605}^{10} \approx 1.99506308 < {10}^{1}$

So the next decimal place is $\textcolor{red}{0}$

${1.99506308}^{10} \approx 999.002$

is just a shade under ${10}^{3}$, so a good approximation for the next digit is: $\textcolor{red}{3}$

So $\log \left(2\right) \approx 0.30103$