Question

Matemática: a) log √12 sendo log 2= 0,3010 e log 3= ,4771

Answer 1

$\log( \sqrt{12})=\log( \sqrt{2^2\cdot3 } )\\ \log( \sqrt{12})=\log(2\cdot\sqrt{3})\\ \log( \sqrt{12})=\log(2)+\log( \sqrt{3})\\ \log( \sqrt{12})=\log(2)+\log(3)^{ \tfrac{1}{2} }\\ \log( \sqrt{12})=\log(2)+ \dfrac{1}{2}\cdot \log(3)\\ \log( \sqrt{12})=0,3010+ \dfrac{1}{2}\cdot0,4771\\ \log( \sqrt{12})=0,3010+0,2355\\\\ \large\boxed{\log( \sqrt{12})=0,5395}$