Question

How do you evaluate `log 1125`?

Answer 1

`log(1125) = 2log(3)+3-3log(2) ~~ 3.0511525225`

Explanation:

It depends where you are starting from.

Suppose you know:

`color(white)(XX)log(2) ~~ 0.30102999566`

`color(white)(XX)log(3) ~~ 0.47712125472`

(See: http://socratic.org/questions/what-is-the-base-10-logarithm-of-2)

Then you can calculate a good approximation for `log(1125)`

`1125 = 3^2 * 5^3 = 3^2 * (10/2)^3 = (3^2 * 10^3) / 2^3`

So:

`log(1125) = log((3^2 * 10^3) / 2^3)`

`= log(3^2) + log(10^3) - log(2^3)`

`= 2log(3) + 3log(10) - 3log(2)`

`= 2log(3) + 3 - 3log(2)`

`~~ 2*0.47712125472 + 3 - 3*0.30102999566`

`~~ 3.0511525225`