Question

Before the invention of calculators, logarithms took much longer to evaluate. (Think of the older generations who went to college in the 1920s... The math they learned was the same, but was all done BY HAND). In the back of their textbooks, they had appendices that looked like this:

? ????	ln(? ????)	log(? ????)
2	ln(2) = 0.6931	log(2) = 0.3010
3	ln(3) = 1.0986	log(3) = 0.4771
5	ln(5) = 1.6094	log(5) = 0.6990
7	ln(7) = 1.9460	log(7) = 0.8451
11	ln(11) = 2.3979	log(11) = 1.0414

Using the fact that the number 679, 140 = 2^2 ∗ 3^2 ∗ 5 ∗ 7^3 ∗ 11 as its prime factorization: (a) use the properties of logs to expand ??(679, 140) as far as possible Hint... expand

ln(2^2∗ 3^2 ∗ 5 ∗ 7^3 ∗ 11) as far as possible
(b) using the table above, plug in the values for each logarithm to four decimal places. What is your answer

using the rounded values in the table?

(c) Using your calculator, what is the value of ??(679140) to 6 decimal places? Explain your answer and why it is similar/different than the answer in part (b)

Answer 1

Answee a)

Properties of log

ln(ab) = ln(a) + ln(b)

ln(aⁿ) = n ln(a)

ln(679140) = ln(2²x3²x5x7³x11)

= ln(2²) + ln(3²) + ln(5) + ln(7³) + ln(11)  

= 2 ln(2) + 2 ln(3) + ln(5) + 3 ln(7) + ln(11)

Answer b)

= 2 (0.6931) + 2 (1.0986) + 1.6094 + 3 (1.9460) + 2.3979

= 13.3087

Answer c)

The value of ln(679140) from calculator is 13.428582

The answer of b and c parts are different because in b part the values put are rounded off values or we leave the last digits thats why both values are different.