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

See 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*f(x=2)+f(x=1.25))=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##)