# How do you calculate ln(1.76)?

Aug 22, 2017

$\ln 1.76 = {\int}_{1}^{1.76} \frac{1}{t} \mathrm{dt} \approx 0.565$

#### Explanation:

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

We could try using the series:

$\ln \left(1 + t\right) = t - {t}^{2} / 2 + {t}^{3} / 3 - {t}^{4} / 4 + \ldots$

but it would converge somewhat slowly.

We could evaluate the integral:

${\int}_{1}^{1.76} \frac{1}{t} \mathrm{dt}$

using one of the popular approximation methods.

For example, Simpson's rule tells us that:

${\int}_{a}^{b} f \left(t\right) \mathrm{dt} \approx \frac{b - a}{6} \left(f \left(a\right) + 4 f \left(\frac{a + b}{2}\right) + f \left(b\right)\right)$

So putting $f \left(t\right) = \frac{1}{t}$, $a = 1$ and $b = 1.76$, we find:

$\ln 1.76 = {\int}_{1}^{1.76} \frac{1}{t} \mathrm{dt}$

$\textcolor{w h i t e}{\ln 1.76} \approx \frac{1.76 - 1}{6} \left(\frac{1}{1} + \frac{4}{\frac{1 + 1.76}{2}} + \frac{1}{1.76}\right)$

$\textcolor{w h i t e}{\ln 1.76} \approx \frac{0.76}{6} \left(1 + \frac{8}{2.76} + \frac{1}{1.76}\right)$

$\textcolor{w h i t e}{\ln 1.76} \approx 0.565$