# How do you evaluate log 1125?

Sep 10, 2015

$\log \left(1125\right) = 2 \log \left(3\right) + 3 - 3 \log \left(2\right) \approx 3.0511525225$

#### Explanation:

It depends where you are starting from.

Suppose you know:

$\textcolor{w h i t e}{X X} \log \left(2\right) \approx 0.30102999566$

$\textcolor{w h i t e}{X X} \log \left(3\right) \approx 0.47712125472$

Then you can calculate a good approximation for $\log \left(1125\right)$

$1125 = {3}^{2} \cdot {5}^{3} = {3}^{2} \cdot {\left(\frac{10}{2}\right)}^{3} = \frac{{3}^{2} \cdot {10}^{3}}{2} ^ 3$

So:

$\log \left(1125\right) = \log \left(\frac{{3}^{2} \cdot {10}^{3}}{2} ^ 3\right)$

$= \log \left({3}^{2}\right) + \log \left({10}^{3}\right) - \log \left({2}^{3}\right)$

$= 2 \log \left(3\right) + 3 \log \left(10\right) - 3 \log \left(2\right)$

$= 2 \log \left(3\right) + 3 - 3 \log \left(2\right)$

$\approx 2 \cdot 0.47712125472 + 3 - 3 \cdot 0.30102999566$

$\approx 3.0511525225$