Determine o valor de:

a) $log\sqrt{2} ^{4}$

b)log $\sqrt[3]{4} ^2$

c) $log_{7} \sqrt[5]{7^{2} }$

Nasgovaskov: Na A) é log √2⁴ ?

maridadojhope: simm

Respostas

respondido por: Nasgovaskov

Devemos determinar o valor dos logaritmos.

Mas antes, veja algumas das propriedades dos logaritmos usadas aqui:

i ) quando tem-se um expoente no logaritmando, implica que ele pode ser passado multiplicando:

log (aᵇ) ⇔ b * log (a)

ii ) quando a base e o logaritmando são iguais, implica que é igual a 1:

logₐ (a) ⇔ 1

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Letra A)

$\begin{array}{l}\sf log~(\sqrt{2^4})=log~(\sqrt{16})\\\\\sf log~(\sqrt{2^4})=log~(4)\\\\\sf log~(\sqrt{2^4})=log~(2^2)\\\\\sf log~(\sqrt{2^4})=2\cdot log~(2)\\\\\!\boxed{\sf log~(\sqrt{2^4})=2log~(2)}\end{array}$

Ou ainda, como é sabido que log (2) ≈ 0,301:

$\begin{array}{l}\sf log~(\sqrt{2^4})=2\cdot0,301\\\\\!\boxed{\sf log~(\sqrt{2^4})\,\approx\,0,602}\end{array}$

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Letra B)

$\begin{array}{l}\sf log~(\sqrt[\sf3]{\sf4^2})=log~(4^{\frac{2}{3}})\\\\\sf log~(\sqrt[\sf3]{\sf4^2})=log~((2^2)^{\frac{2}{3}})\\\\\sf log~(\sqrt[\sf3]{\sf4^2})=log~(2^{\frac{4}{3}})\\\\\!\boxed{\sf log~(\sqrt[\sf3]{\sf4^2})=\dfrac{4}{3}\cdot log~(2)}\end{array}$

Ou ainda, como é sabido que log (2) ≈ 0,301:

$\begin{array}{l}\sf log~(\sqrt[\sf3]{\sf4^2})=\dfrac{4}{3}\cdot0,301\\\\\sf log~(\sqrt[\sf3]{\sf4^2})=\dfrac{1,204}{3}\\\\\!\boxed{\sf log~(\sqrt[\sf3]{\sf4^2})\,\approx\,0,401}\end{array}$

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Letra C)

$\begin{array}{l}\sf log_{\:7}~(\sqrt[\sf5]{\sf7^2})=log_{\:7}~(7^{\frac{2}{5}})\\\\\sf log_{\:7}~(\sqrt[\sf5]{\sf7^2})=\dfrac{2}{5}\cdot log_{\:7}~(7)\\\\\sf log_{\:7}~(\sqrt[\sf5]{\sf7^2})=\dfrac{2}{5}\cdot1\\\\\sf log_{\:7}~(\sqrt[\sf5]{\sf7^2})=\dfrac{2}{5}\\\\\!\boxed{\sf log_{\:7}~(\sqrt[\sf5]{\sf7^2})=0,4}\end{array}$