How do you calculate #ln(1.76)#?

1 Answer
Aug 22, 2017

Answer:

#ln 1.76 = int_1^1.76 1/t dt ~~ 0.565#

Explanation:

Hmmm. I'm not sure exactly what we are looking for here.

We could try using the series:

#ln(1+t) = t-t^2/2+t^3/3-t^4/4+...#

but it would converge somewhat slowly.

We could evaluate the integral:

#int_1^1.76 1/t dt#

using one of the popular approximation methods.

For example, Simpson's rule tells us that:

#int_a^b f(t) dt ~~ (b-a)/6 (f(a)+4f((a+b)/2)+f(b))#

So putting #f(t) = 1/t#, #a=1# and #b=1.76#, we find:

#ln 1.76 = int_1^1.76 1/t dt#

#color(white)(ln 1.76) ~~ (1.76-1)/6(1/1+4/((1+1.76)/2)+1/1.76)#

#color(white)(ln 1.76) ~~ 0.76/6(1+8/2.76+1/1.76)#

#color(white)(ln 1.76) ~~ 0.565#