Question

questão 54 por favor S=(1/4)

Answer 1

Propriedade:

$\boxed{\boxed{log_{b}(a^{n})=n\cdot log_{b}(a)}}$

Mudança de base (b para c):

$\boxed{\boxed{log_{b}(a)=\dfrac{log_{c}(a)}{log_{c}(b)}}}$
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$log_{3}(5)\cdot log_{4}(27)\cdot log_{25}(\sqrt[3]{2})$

Mudando as bases para 10:

$log_{3}(5)\cdot log_{4}(27)\cdot log_{25}(\sqrt[3]{2})=\dfrac{log(5)}{log(3)}\cdot\dfrac{log(27)}{log(4)}\cdot\dfrac{log(\sqrt[3]{2})}{log(25)}\\\\\\log_{3}(5)\cdot log_{4}(27)\cdot log_{25}(\sqrt[3]{2})=\dfrac{log(5)}{log(3)}\cdot\dfrac{log(3^{3})}{log(2^{2})}\cdot\dfrac{log(\sqrt[3]{2^{1}})}{log(5^{2})}\\\\\\log_{3}(5)\cdot log_{4}(27)\cdot log_{25}(\sqrt[3]{2})=\dfrac{log(5)}{log(3)}\cdot\dfrac{3\cdot log(3)}{2\cdot log(2)}\cdot\dfrac{log(2^{1/3})}{2\cdot log(5)}$

Cortando log(3) e log(5):

$log_{3}(5)\cdot log_{4}(27)\cdot log_{25}(\sqrt[3]{2})=\dfrac{3}{2\cdot log(2)}\cdot\dfrac{(\frac{1}{3})log(2)}{2}\\\\\\log_{3}(5)\cdot log_{4}(27)\cdot log_{25}(\sqrt[3]{2})=3\cdot\dfrac{1}{3}\cdot\dfrac{1}{2}\cdot\dfrac{1}{2}\\\\\\\boxed{\boxed{log_{3}(5)\cdot log_{4}(27)\cdot log_{25}(\sqrt[3]{2})=\dfrac{1}{4}}}$