# How do you calculate log 1300?

May 20, 2017

$\log 1300 \approx 3.1139434$

#### Explanation:

Suppose you know:

$\log 2 \approx 0.30103$

$\ln 10 \approx 2.302585$

Then:

$1300 = 10 \cdot 2 \cdot 65 = 10 \cdot 2 \cdot 64 \cdot \frac{65}{64} = 10 \cdot {2}^{7} \cdot \left(1 + \frac{1}{64}\right)$

So:

$\log 1300 = \log \left(10 \cdot {2}^{7} \cdot \left(1 + \frac{1}{64}\right)\right)$

$\textcolor{w h i t e}{\log 1300} = \log 10 + 7 \log 2 + \log \left(1 + \frac{1}{64}\right)$

$\textcolor{w h i t e}{\log 1300} = \log 10 + 7 \log 2 + \ln \frac{1 + \frac{1}{64}}{\ln} 10$

Now:

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

So:

$\ln \left(1 + \frac{1}{64}\right) = \frac{1}{64} - \frac{1}{2 \cdot {64}^{2}} + \frac{1}{3 \cdot {64}^{3}} - \frac{1}{4 \cdot {64}^{4}} + \ldots$

$\textcolor{w h i t e}{\ln \left(1 + \frac{1}{64}\right)} = \frac{1}{64} - \frac{1}{8192} + \frac{1}{786432} - \frac{1}{67108864} + \ldots$

$\textcolor{w h i t e}{\ln \left(1 + \frac{1}{64}\right)} \approx 0.01550419$

and:

$\log 1300 = \log 10 + 7 \log 2 + \ln \frac{1 + \frac{1}{64}}{\ln} 10$

$\textcolor{w h i t e}{\log 1300} \approx 1 + 7 \cdot 0.30103 + \frac{0.01550419}{2.302585}$

$\textcolor{w h i t e}{\log 1300} \approx 3.1139434$