Question

Considerando log 2 = 0,301 , log 3 = 0,4771 , log 7 = 0,845

a) log $\sqrt[3]{60}$
​

Answer 1

log∛60 = $log(2^2.3.5)^{1/3}$

$log(2^2.3.5)^{1/3} = \frac{1}{3}log 2^2.3.5$

$\frac{2}{3} log2 + \frac{1}{3} log3 + \frac{1}{3} log 5$

$\frac{2}{3}.0,301 + \frac{1}{3}0,4771 + \frac{1}{3} log5$

$\frac{2}{3}.0,301 + \frac{1}{3}0,4771 + \frac{1}{3} log\frac{10}{2}$

0,20066... + 0,159033... + $\frac{1}{3} (log10 - log2)$

$0,359693...+$ $\frac{1}{3} -\frac{1}{3}.0,301$

0,693026 - 0,10033...

≅ 0,592696