Question

How do you evaluate `ln1.4`?

Answer 1

Use a calculator or Taylor series to find:

`ln(1.4) ~~ 0.3364722`

Explanation:

My calculator tells me that:

`ln(1.4) ~~ 0.33647223662121293050`

but can we calculate it without a `ln` function?

We can use the Taylor series:

`ln(1+x) = sum_(n=1)^oo (-1)^(n-1) x^n/n`

So:

`ln(1.4) = ln(1+0.4)`

`= 0.4-0.4^2/2+0.4^3/3-0.4^4/4+...`

`~~0.4-0.08+0.02133333-0.0064+0.002048-0.00068267+0.00023406-0.00008192+0.00002913-0.00001049+0.00000381-0.00000140+0.00000052-0.00000019+0.00000007-0.00000003+0.00000001 = 0.33647223`