How do you evaluate $log 1125$ ?

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Answer 1 · 2015-09-10T20:25:38

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

It depends where you are starting from.

Suppose you know:

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

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

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:

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

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

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

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

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

$~~ 3.0511525225$

How do you evaluate log 1125log 1125?