Question

considerando as aproximações log2=0,3 e log3=0,48,calcule log da raiz qualdrada de 24

Answer 1

Olá,

use as propriedades,

Produto:

$\mathsf{log_b(a\cdot c)=log_b(a)+log_b(c)}$

Potência:

$\mathsf{ log_b(a)^n=n\cdot log_b(a)}$

........................

$\mathsf{log( \sqrt{24})=log( \sqrt{3\cdot2^3})}\\ \mathsf{log( \sqrt{24})=log(3\cdot2^3)^{ \tfrac{1}{2} }}\\\\ \mathsf{log( \sqrt{24})= \dfrac{1}{2} \cdot log(3\cdot2^3)}\\\\ \mathsf{log( \sqrt{24})= \dfrac{1}{2}\cdot[log(3)+log(2)^3] }\\\\ \mathsf{log( \sqrt{24})= \dfrac{1}{2}\cdot[log(3)+3\cdot log(2)] }\\\\ \mathsf{log( \sqrt{24})= \dfrac{1}{2}\cdot[0,48+3\cdot0,3] }\\\\ \mathsf{log( \sqrt{24})= \dfrac{1}{2}\cdot[0,48+0,9] }\\\\ \mathsf{log( \sqrt{24})= \dfrac{1}{2}\cdot1,38 }$


$\huge\boxed{\mathsf{log( \sqrt{24})=0,69}}$

Tenha ótimos estudos ;D