Question

Dados log 2 = 0,301 e log 3 = 0,477, calcule:

log 125

log 3600

log 0,1008

log (2√3)

log (3∛4)

(Necessito das resoluções)

Answer 1

$1)\\\\ log125=log5^3=\\\\ =3.log5=3.(log\frac{10}{2})=\\\\ =3(log10-log2)=3.(1-0,301)=\\\\ =3.(0,699)=2,097\\\\\\ 2)\\\\ log3600=log(36.100)=log36+log100=\\\\ =log6^2+log100=2.log6+log100=\\\\ =2(log2.3)+log100=\\\\ =2(log2+log3)+log100=\\\\ =2(0,301+0,477)+2=\\\\ =2.(0,778)+2=1,556+2=3,556$

$3)\\\\Penso \ que \ o \ correto \ \'e \ 0,108\\\\ log0,108=log\frac{108}{1000}=\\\\ =log108-log1000=log(4.27)-log1000=\\ =log4+log27-log1000=\\ =log2^2+log3^3-log1000=\\ =2log2+3log3-log1000\\ =2.(0,301)+3.(0,477)-3=\\ =0,602+1,431-3=-0,967\\\\ 4)\\\\ log2\sqrt{3}=log2+log\sqrt{3}=\\\\ =log2+log3^{\frac{1}{3}}=\\\\ =log2+\frac{1}{3}.log3=\\\\ =0,301+\frac{1}{3}.0,477=0,301+0,159=0,46$

$5)\\\\ log3 \sqrt[3]{4} =log3+log \sqrt[3]{4}=\\\\ =log3+log \sqrt[3]{2^2}=\\\\ =log3+log2^{\frac{2}{3}}=\\\\ =log3+\frac{2}{3}.log2=\\\\ =0,477+\frac{2}{3}.(0,301)=\\\\ =0,477+0,20066=0,6776...$