How do I solve 'log(base 10) 5' without using the calculator?

1 Answer
Apr 16, 2016

Answer:

See explanation

Explanation:

If you have memorized that
#log2=0.3#
you can follow this way
#log5=log(10/2)=1-log2=1-0.3=0.7#

If you want a general way to find logarithms without using calculators or tables, you could use this formula:
#(1/2)ln|(1+x)/(1-x)|=f(x)=x+x^3/3+x^5/5+...#
And
#logy=lny/ln10=2/ln10*(1/2*ln|y|)# => #logy=0.869*(1/2*ln|y|)# where #y=(1+x)/(1-x)#
(Note1: you can use #2/ln10= 0.868589# with the precision you like. Using two terms of the series, 0.869 has a proper level of precision. Note 2: the values of x must be smaller than 1.)

We can't calculate #log5# directly because
#(x+1)/(1-x)=5# => #x+1=5-5x# => #6x=4# => #x=1.5#
And the series doesn't converge when #x>1#

But since #5=2*2.5#
for #y_1=2 ->(x+1)/(1-x)=2# => #x+1=2-2x# => #x=1/3~=0.3333#
#f(x=1/3)=1/3+1/3^3*1/3=1/3+1/81=0.3333+0.0123=0.3456#

for #y_2=2.5 -> (x+1)/(1-x)=2.5# => #x+1=2.5-2.5x# => #3.5x=1.5# => #x=3/7~=0.4286#
Of course we can use this #x=0.4286#. But perhaps there is an easier way (without a calculator we need to think of this) such as:

Considering that #5=2^2*1.25# (and since we have already calculated #f(x=1/3)#):
for #y_2=1.25 -> (x+1)/(1-x)=1.25# => #x+1=1.25-1.25x# => #2.25x=0.25# => #x=25/225=1/9~=0.1111#
#f(x=1/9)=0.1111+1/9^3*1/3=0.1111+1/729*1/3=1/9+1/2187=0.1111+0.0005=0.1116#
(as to the number #0.0005# just remember that #10/2=5#)

Using the results above
#log5=0.869[2*(1/2*ln|2|)+(1/2*ln|1.25|)]=0.869[2*f(x=1/3)+f(x=1/9)]=0.869[2*0.3456+0.1116]=0.869[0.6912+0.1116]=0.869*0.8028=0.6976332# or #0.698# in 3 decimals

We should be aware that this last estimate is smaller than the correct result.

(In fact #log5=0.6990#)