Question

1- sabendo que log2=0,301 ,log3=0,477 ,log5=0,699 ,calcule
a) log∛5
b) log $\sqrt[5]{6}$

Answer 1

a)

$log5^{1/3}= \frac{1}{3}log5= \frac{1}{3}.0,669=0,223$
b)

$log6^{1/5}=log(2.3)^{1/5}=log2^{1/5}+log3^{1/5}= \frac{1}{5}log2+ \frac{1}{5}log3= \\ \\ \frac{1}{5}.0,301+ \frac{1}{5}.0,447= \frac{0,301+0,447}{5}= \frac{0,748}{5}=0,1496$