# How can I calculate a logarithm without a calculator?

Oct 30, 2015

See explanation, where I show how to find ${\log}_{2} \left(7\right) \approx 2.8$

#### Explanation:

Since you were not specific as to what base of logarithm you wanted, I will take the liberty of showing you how to calculate logarithms base 2 in binary. They are perhaps the easiest to do by hand. Elsewhere I have shown a method to calculate common (base $10$) logarithms, but that involves a lot of raising numbers to the $10$th power so is rather tedious.

Let us calculate ${\log}_{2} 7$.

From now on express numbers in binary...

We want to calculate ${\log}_{{10}_{2}} {111}_{2}$

${10}_{2}^{{10}_{2}} = {100}_{2} < {111}_{2} < {1000}_{2} = {10}_{2}^{{11}_{2}}$

So the digits before the binary point are $\textcolor{b l u e}{10}$

Next divide ${111}_{2}$ by ${10}_{2}^{{10}_{2}} = {100}_{2}$ to get ${1.11}_{2}$

Square ${1.11}_{2}$ to get ${11.0001}_{2}$

Since this is greater than ${10}_{2}$, the first digit after the binary point is $\textcolor{b l u e}{1}$

Divide by ${10}_{2}$ to get ${1.10001}_{2}$

Square ${1.10001}_{2}$ to get ${10.0101100001}_{2}$

Since this is greater than ${10}_{2}$, the second digit after the binary point is $\textcolor{b l u e}{1}$

Divide by ${10}_{2}$ to get ${1.00101100001}_{2}$

Square ${1.00101100001}_{2}$ to get ${1.0101111111011011000001}_{2}$

Since this is less than ${10}_{2}$, the third digit after the binary point is $\textcolor{b l u e}{0}$

Square ${1.0101111111011011000001}_{2}$ to get ${1.11100011100110100101000001010111110110000001}_{2}$

Since this is less than ${10}_{2}$, the fourth digit after the binary point is $\textcolor{b l u e}{0}$

To cut down on the arithmetic, I will approximate this as ${1.11100011100110100101}_{2}$

Square ${1.11100011100110100101}_{2}$ to get ${11.1001000110001111101001101110010001011001}_{2}$

Since this is greater than ${10}_{2}$, the fifth digit after the binary point is $\textcolor{b l u e}{1}$

Divide by ${10}_{2}$ to get ${1.11001000110001111101001101110010001011001}_{2}$

I'll stop here, but I hope you get the idea.

Putting the digits we have found together we get:

${\log}_{{10}_{2}} {111}_{2} \approx {10.11001}_{2} = 2 + \frac{25}{32} \approx 2.8$

Actually ${\log}_{2} \left(7\right) \approx 2.80735$