# How can you calculate the value of log(0.9863) ?

Aug 19, 2017

$\log \left(0.9863\right) \approx - 0.005991$

#### Explanation:

Suppose we know a suitable approximation for $\ln 10$, say:

$\ln 10 \approx 2.302585$

There is a series for $\ln \left(1 + x\right)$ like this:

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

So putting $x = 0.9863 - 1 = - 0.0137$, we have:

$\ln \left(0.9863\right) = \left(- 0.0137\right) - {\left(- 0.0137\right)}^{2} / 2 + {\left(- 0.0137\right)}^{3} / 3 - {\left(- 0.0137\right)}^{4} / 4 + \ldots$

$\textcolor{w h i t e}{\ln \left(0.9863\right)} = - 0.0137 - \frac{0.00018769}{2} - \frac{0.000002571353}{3} - \ldots$

$\textcolor{w h i t e}{\ln \left(0.9863\right)} = - 0.0137 - 0.000093845 - 0.000000857117 \overline{6} - \ldots$

$\textcolor{w h i t e}{\ln \left(0.9863\right)} \approx - 0.013794702$

Then by the change of base formula:

$\log \left(0.9863\right) = \ln \frac{0.9863}{\ln} \left(10\right) \approx - \frac{0.013794702}{2.302585} \approx - 0.00599096$

Since the argument $0.9863$ was only quoted to $4$ significant digits, we should probably limit the precision of our answer and say:

$\log \left(0.9863\right) \approx - 0.005991$