Question

Resolva os logaritmos abaixo:
a) log16 64
b) log5 0,000064
c) log49 √73
d) log√2 4
e) log √43 2
f) log2√8
g) log3 27
h) log4 8
i) log2 0,25
j) log2√64

Answer 1

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EXPLICAÇÃO PASSO-A-PASSO______✍

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☺lá, Gabi, como tens passado nestes tempos de quarentena⁉ E os estudos à distância, como vão⁉ Espero que bem❗ Acompanhe a resolução abaixo. ✌

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Ⓐ_____________________________✍

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$\huge\gray{\boxed{\sf\blue{ log_{16}(64) }}}$

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➡ $\large\blue{\sf 16^x = 64 }$

➡ $\large\blue{\sf (4^2)^x = 4^3 }$

➡ $\large\blue{\sf 4^{2x} = 4^3 }$

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☔ Pela simetria e por ambos os lados serem funções bijetivas temos que 2x = 3, mas vamos (pelo menos para esse primeiro exercício) demonstrar a manipulação algébrica que confirma isto. Aplicando a função Logaritmica em ambos os lados da igualdade temos

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➡ $\large\blue{\sf log(4^{2x}) = log(4^3) }$

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☔ Temos como propriedade da Função Logaritmo que uma potência do logaritmando pode ser reescrita como um coeficiente que multiplica o log , conhecida como Regra do Tombo

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$\large\red{\boxed{\pink{\boxed{\begin{array}{rcl}&&\\&\orange{\rm log_c(a^b) \iff b \cdot log_c(a)}&\\&&\\\end{array}}}}}$

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☔ Portanto temos que

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➡ $\large\blue{\sf 2x \cdot log(4) = 3 \cdot log(4) }$

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☔ Dividindo ambos os lados da igualdade por log(4) temos

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➡ $\large\blue{\sf \dfrac{2x \cdot log(4)}{log(4)} = \dfrac{3 \cdot log(4)}{log(4)} }$

➡ $\large\blue{\sf 2x = 3 }$

➡ $\huge\blue{\sf x = 1,5 }$  ✅

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Ⓑ_____________________________✍

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$\huge\gray{\boxed{\sf\blue{ log_{5}(0,000064) }}}$

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➡ $\large\blue{\sf 5^x = \dfrac{2^6}{10^6} }$

➡ $\large\blue{\sf 5^x = \dfrac{2^6}{2 \cdot 5)^6} }$

➡ $\large\blue{\text{ \sf 5^x = \dfrac{\diagup\!\!\!\!{2}^6}{\diagup\!\!\!\!{2}^6 \cdot 5^6} }}$

➡ $\large\blue{\sf 5^x = \dfrac{1}{5^6} }$

➡ $\large\blue{\sf 5^x = 5^{-6} }$

➡ $\huge\blue{\sf x = -6 }$  ✅

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Ⓒ_____________________________✍

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$\huge\gray{\boxed{\sf\blue{ log_{49}(\sqrt{73}) }}}$

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➡ $\large\blue{\text{ \sf 49^x = 73^{\frac{1}{2}} }}$

➡ $\large\blue{\text{ \sf log(49^x) = log(73^{\frac{1}{2}}) }}$

➡ $\large\blue{\sf x \cdot log(49) = \dfrac{log(73)}{2} }$

➡ $\huge\blue{\sf x = \dfrac{log(73)}{2 \cdot log(49)} }$  ✅

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Ⓓ_____________________________✍

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$\huge\gray{\boxed{\sf\blue{ log_{\sqrt{2}}(4) }}}$

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➡ $\large\blue{\text{ \sf (2^{\frac{1}{2}})^x = 2^2 }}$

➡ $\large\blue{\text{ \sf 2^{\frac{x}{2}} = 2^2 }}$

➡ $\large\blue{\sf \dfrac{x}{2} = 2 }$

➡ $\huge\blue{\sf x = 4 }$  ✅

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Ⓔ_____________________________✍

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$\huge\gray{\boxed{\sf\blue{ log_{\sqrt{43}}(2) }}}$

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➡ $\large\blue{\text{ \sf (43^{\frac{1}{2}})^x = 2 }}$

➡ $\large\blue{\text{ \sf 43^{\frac{x}{2}} = 2 }}$

➡ $\large\blue{\text{ \sf log(43^{\frac{x}{2}}) = log(2) }}$

➡ $\large\blue{\sf \dfrac{x \cdot log(43)}{2} = log(2) }$

➡ $\huge\blue{\sf x = \dfrac{2 \cdot log(2)}{log(43)} }$  ✅

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Ⓕ_____________________________✍

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$\huge\gray{\boxed{\sf\blue{ log_{2}(8) }}}$

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➡ $\large\blue{\sf 2^x = 2^3 }$

➡ $\huge\blue{\sf x = 3 }$  ✅

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Ⓖ_____________________________✍

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$\huge\gray{\boxed{\sf\blue{ log_{3}(27) }}}$

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➡ $\large\blue{\sf 3^x = 3^3 }$

➡ $\huge\blue{\sf x = 3 }$  ✅

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Ⓗ_____________________________✍

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$\huge\gray{\boxed{\sf\blue{ log_{4}(8) }}}$

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➡ $\large\blue{\sf (2^2)^x = 2^3 }$

➡ $\large\blue{\sf 2^{2x} = 2^3 }$

➡ $\large\blue{\sf 2x = 3 }$

➡ $\huge\blue{\sf x = 1,5 }$  ✅

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Ⓘ_____________________________✍

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$\huge\gray{\boxed{\sf\blue{ log_{2}(0,25) }}}$

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➡ $\large\blue{\sf 2^x = \dfrac{5^2}{10^2} }$

➡ $\large\blue{\sf 2^x = \dfrac{5^2}{(2 \cdot 5)^2} }$

➡ $\large\blue{\text{ \sf 2^x = \dfrac{\diagup\!\!\!\!{5}^2}{2^2 \cdot \diagup\!\!\!\!{5}^2} }}$

➡ $\large\blue{\sf 2^x = \dfrac{1}{2^2} }$

➡ $\large\blue{\sf 2^x = 2^{-2} }$

➡ $\huge\blue{\sf x = -2 }$  ✅

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Ⓙ_____________________________✍

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$\huge\gray{\boxed{\sf\blue{ log_{2}(\sqrt{64}) }}}$

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➡ $\large\blue{\text{ \sf 2^x = (2^6)^{\frac{1}{2}} }}$

➡ $\large\blue{\text{ \sf 2^x = 2^{\frac{6}{2}} }}$

➡ $\large\blue{\sf 2^x = 2^3 }$

➡ $\huge\blue{\sf x = 3 }$  ✅

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_______________________________☁

☕ Bons estudos.

(Dúvidas nos comentários) ☄

__________________________$\LaTeX$✍

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