Question

URGENTE, VALENDO 50 PORQUE PRECISO MUITO.
Dados log2= 0,301, log3= 0,477 e log7=0845, calcule.

A) Log ⁴√125

B) Log 25/6

C) Log 35

Answer 1

No enunciado deveria ter o log de 5 também. As respostas estão na foto.

Answer 2

Ae mano,

Use as propriedades,

Produto:

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

Quociente:

$\mathsf{log_b\left(\dfrac{a}{c}\right)=log_b(a)-log_b(c) }$

Potência:

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

Decorrente da Definição:

$\mathsf{log_b(b)=1~~(No~caso,~a~base~10)\Rightarrow log(10)\Rightarrow log_{10}(10)=1}}$

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$\mathsf{log(^4 \sqrt{125})=log(^4 \sqrt{5^3}) }\\ \mathsf{log(^4 \sqrt{125})=log(5^{ \tfrac{3}{4} }) }\\\\ \mathsf{log(^4 \sqrt{125})= \dfrac{3}{4}\cdot log(5) }\\\\ \mathsf{log(^4 \sqrt{125})= \dfrac{3}{4}\cdot log\left( \dfrac{10}{2} \right) }\\\\ \mathsf{log(^4 \sqrt{125})= \dfrac{3}{4}\cdot[log(10)-log(2)] }\\\\ \mathsf{log(^4 \sqrt{125})= \dfrac{3}{4}\cdot[log_{10}(10)-log(2)] }\\\\ \mathsf{log(^4 \sqrt{125})= \dfrac{3}{4}\cdot(1-0,301) }$

$\mathsf{log(^4 \sqrt{125})= \dfrac{3}{4}\cdot0,699 }\\\\ \Large\boxed{\mathsf{log(^4 \sqrt{125})\approx0,5242} }$

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

$\mathsf{log\left( \dfrac{25}{6} \right)=log\left( \dfrac{5^2}{2\cdot3}\right) }\\\\ \mathsf{log\left( \dfrac{25}{6} \right)=[log(5^2)]-log(2\cdot3)}\\\\ \mathsf{log\left( \dfrac{25}{6}\right)=[log(5)^2]-log(2)+log(3)}\\\\ \mathsf{log\left( \dfrac{25}{6}\right)=\{[2\cdot log(5)]-[log(2)+log(3)]}\\\\ \mathsf{log\left( \dfrac{25}{6}\right)=\left[2\cdot log\left( \dfrac{10}{2} \right)\right]-[log(2)+log(3)]}$

$\mathsf{log\left( \dfrac{25}{6}\right)=\{2\cdot [log(10)-[log(2)\}-[log(2)+log(3)}\\\\ \mathsf{log\left( \dfrac{25}{6}\right)=\{2\cdot[1-0,301]\}-[0,301+0,477]}\\\\ \mathsf{log\left( \dfrac{25}{6}\right)=[2\cdot0,699]-[0,778]}\\\\ \mathsf{log\left( \dfrac{25}{6}\right)=1,398-0,778}\\\\ \Large\boxed{\mathsf{log\left( \dfrac{25}{6}\right)=0,62}}$

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$\mathsf{log(35)=log(5\cdot7)}\\ \mathsf{log(35)=log(5)+log(7)}\\\\ \mathsf{log(35)=log\left( \dfrac{10}{2}\right)+log(7)}\\\\\mathsf{log(35)=[log(10)-log(2)]+log(7)}\\\\\mathsf{log(35)=[1-0,301]+0,845}\\\\ \Large\boxed{\mathsf{log(35)=1,544}}$

Tenha ótimos estudos mano!