How do you evaluate $log 1125$ $log 1125$ ?

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How do you evaluate $log 1125$ $log 1125$ ?

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Answer 1 · 2015-09-10T20:25:38

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

Explanation:

It depends where you are starting from.

Suppose you know:

$X X log (2) \approx 0.30102999566$ $color(white)(XX)log(2) ~~ 0.30102999566$

$X X log (3) \approx 0.47712125472$ $color(white)(XX)log(3) ~~ 0.47712125472$

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

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

$1125 = 3^{2} \cdot 5^{3} = 3^{2} \cdot {(\frac{10}{2})}^{3} = \frac{3^{2} \cdot 10^{3}}{2^{3}}$ $1125 = 3^2 * 5^3 = 3^2 * (10/2)^3 = (3^2 * 10^3) / 2^3$

So:

$log (1125) = log (\frac{3^{2} \cdot 10^{3}}{2^{3}})$ $log(1125) = log((3^2 * 10^3) / 2^3)$

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

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

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

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

$\approx 3.0511525225$ $~~ 3.0511525225$

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How do you evaluate $log 1125$ $log 1125$ ?

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