Calcule o valor de cada expressão:

4log8 2√2 + log16 0,04

6^(2+log6 7) − log(log10)

log3 raiz quarta de 9 − 4^log4 5

2log9 1+[log0,5 (1/8)]^2

Respostas

respondido por: Nasgovaskov

Antes de tudo, confira algumas das propriedades dos logaritmos usadas para as resoluções:

logₐ (bᶜ) ⇔ c * logₐ (b)

o expoente do logaritmando passa multiplicando o logaritmo

logₐᶜ (b) ⇔ 1/c * logₐ (b)

o expoente da base passa sendo o seu inverso multiplicando o logaritmo

logₐ (b * c) ⇔ logₐ (b) + logₐ (c)

o produto de fatores no logaritmando passa a ser uma soma de logaritmos de cada fator

a^(logₐ (b)) ⇔ b

se a base da potência e a base do logaritmo (sendo o log como expoente) passa a ser somente o logaritmando

logₐ (1) ⇔ 0

o logaritmo de um em qualquer base, é zero

logₐ (a) ⇔ 1

quando a base e o logaritmando são iguais, é um

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*obs.: lembre-se que quando a base não aparece, é igual a 10: log (a) = log₁₀ (a)

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Agora vamos calcular o valor das expressões.

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( I )

$\begin{array}{l}\sf 4log_8~(2\sqrt{2})+log_{16}~(0,04)\\\\\sf log_8~((2\sqrt{2})^4)+log_{2^4}~\bigg(\dfrac{4}{100}\bigg)\\\\\sf log_8~(2^4\cdot2^2)+\dfrac{1}{4}\cdot log_2~\bigg(\dfrac{100}{4}^{-1}\bigg)\\\\\sf log_8~(2^3\cdot2^3)+\dfrac{1}{4}\cdot log_2~(25^{-1})\\\\\sf log_8~(8\cdot8)+\dfrac{1}{4}\cdot log_2~(5^{-2})\\\\\sf log_8~(8)+log_8~(8)-2\cdot\dfrac{1}{4}\cdot log_2~(5)\\\\\sf1+1-\dfrac{2}{4}\cdot log_2~(5)\\\\\!\boxed{\sf2-\dfrac{1}{2}log_2~(5)}\end{array}$

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( II )

$\begin{array}{l}\sf6^{2 \: + \: log_6~(7)}-log~(log~(10))\\\\\sf6^2\cdot6^{log_6~(7)}-log_{10}~(log_{10}~(10))\\\\\sf36\cdot7-log_{10}~(1)\\\\\sf252-0\\\\\!\boxed{\sf252}\end{array}$

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( III )

$\begin{array}{l}\sf log_3~(\sqrt[\sf4]{\sf9})-4^{log_4~(5)}\\\\\sf log_3~(9^\frac{1}{4})-5\\\\\sf log_3~((3^2)^\frac{1}{4})-5\\\\\sf log_3~(3^\frac{2}{4})-5\\\\\sf\dfrac{2}{4}\cdot log_3~(3)-5\\\\\sf\dfrac{1}{2}\cdot1-5\\\\\sf\dfrac{-5\cdot2+1}{2}\\\\\!\boxed{\sf-\dfrac{9}{2}}\end{array}$

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( IV )

$\begin{array}{l}\sf2log_9~(1)+\bigg[log_{0,5}~\bigg(\dfrac{1}{8}\bigg)\bigg]^2\\\\\sf2\cdot0+[log_\frac{1}{2}~(8^{-1})]^2\\\\\sf0+[log_{2^{-1}}~((2^3)^{-1})]^2\\\\\sf[log_{2^{-1}}~(2^{-3})]^2\\\\\sf[-\dfrac{1}{1}\cdot(-3)\cdot log_2~(2)]^2\\\\\sf[3\cdot1]^2\\\\\sf3^2\\\\\!\boxed{\sf9}\end{array}$

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Portanto, o resultado final de cada expressão:

$\boxed{\begin{array}{l}\\\sf (\,I\,)~~~4log_8~(2\sqrt{2})+log_{16}~(0,04)=\boldsymbol{\sf2-\dfrac{1}{2}log_2~(5)}\\\\\sf(\,II\,)~~~6^{2+log_6~(7)}-log~(log~(10))=\boldsymbol{\sf252}\\\\\sf(\,III\,)~~~log_3~(\sqrt[\sf4]{\sf9})-4^{log_4~(5)}=\boldsymbol{\sf-\dfrac{9}{2}}\\\\\sf(\,IV\,)~~~2log_9~(1)+\bigg[log_{0,5}~\bigg(\dfrac{1}{8}\bigg)\bigg]^2=\boldsymbol{\sf9}\\\\\end{array}}$