# What is the base-10 logarithm of 2?

##### 2 Answers
May 28, 2015

${\log}_{10} 2 = x$
where ${10}^{x} = 2$

There is no simple way to calculate the value of $x$
but you can look it up (using a calculator, for example)
${\log}_{10} 2 = 0.30103$ (approx.)

May 28, 2015

The base-10 logarithm of 2 is the number $x$ such that ${10}^{x} = 2$.

You can calculate logarithms by hand using just multiplication (and dividing by powers of 10 - which is just digit shifting) and the fact that ${\log}_{10} \left({x}^{10}\right) = 10 \cdot {\log}_{10} x$, though it's not very practical...

It goes something like this:

$2 < 10$ so ${\log}_{10} \left(2\right) < 1$

So the part of ${\log}_{10} \left(2\right)$ before the decimal place is $\ast 0 \ast$.

Calculate ${2}^{10}$ by multiplying it by itself to get:

${2}^{10} = 1024$

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

So $3 = {\log}_{10} \left({10}^{3}\right) < {\log}_{10} \left({2}^{10}\right) < \log \left({10}_{4}\right) = 4$

Now ${\log}_{10} \left({2}^{10}\right) = 10 \cdot {\log}_{10} \left(2\right)$

so by raising to the $10 t h$ power, we have multiplied the logarithm by 10.

So the first digit of ${\log}_{10} \left(2\right)$ after the decimal point is $\ast 3 \ast$.

Then

${\log}_{10} \left({2}^{10}\right) - 3 = {\log}_{10} \left({2}^{10}\right) - {\log}_{10} \left(1000\right)$

$= {\log}_{10} \left({2}^{10} / 1000\right) = {\log}_{10} \left(\frac{1024}{1000}\right) = {\log}_{10} \left(1.024\right)$

So having found the digit $3$, we next divide $1024$ by ${10}^{3}$ to get $1.024$ and repeat the process to get the next digit:

${1.024}^{10} \cong 1.2676506$

This is still less than $10$, so the next digit is $\ast 0 \ast$.

${1.2676506}^{10} \cong 10.715086$

${10}^{1} = 10 \le 10.715086 < 100 = {10}^{2}$, so the next digit is $\ast 1 \ast$.

Divide $10.715086$ by ${10}^{1}$ to get $1.0715086$

${1.0715086}^{10} \cong 1.99506298$

This is less than $10$, so the next digit is $\ast 0 \ast$.

${1.99506298}^{10} \cong 999.0014 \cong 1000 = {10}^{3}$

I'll stop here at this approximation, giving a digit $\ast 3 \ast$.

Collecting our digits, we have ${\log}_{10} \left(2\right) \cong 0.30103$