How can I calculate a logarithm without a calculator?

1 Answer
Oct 30, 2015

Answer:

See explanation, where I show how to find #log_2(7) ~~ 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 #color(blue)(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 #color(blue)(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 #color(blue)(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 #color(blue)(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 #color(blue)(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 #color(blue)(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 ~~ 10.11001_2 = 2+25/32 ~~ 2.8#

Actually #log_2(7) ~~ 2.80735#