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A melhor resposta!
2014-08-23T13:29:41-03:00
E aí mano,

use a definição de logaritmos:

\Large\boxed{log_bc=a~\to~b^{a}=c}

>>>>>>>>>>>>>>>>>>>>>>>>>

log_{ \tfrac{1}{125}} \sqrt[3]{5}=x\\\\\\
 \dfrac{1}{125}^x= \sqrt[3]{5}\\\\
(5^{-3})^x=5^{ \tfrac{1}{3} }\\\\
\not5^{-3x}=\not5^{ \tfrac{1}{3}}\\\\
-3x= \dfrac{1}{3}\\\\
x=- \dfrac{1}{9}

>>>>>>>>>>>>>>>>>>>>>>>>>

log_{ \tfrac{1}{9}}3 \sqrt{3}\\\\\\
 \dfrac{1}{9}^x=3 \sqrt{3}\\\\
(3^{-2})^x= \sqrt{27}\\
\not3^{-2x}=\not3^{ \tfrac{3}{2} }\\\\
-2x= \dfrac{3}{2}\\\\
x=- \dfrac{3}{4}

>>>>>>>>>>>>>>>>>>>>>>>>>

log_x \dfrac{27}{8}=3\\\\
x^3= \dfrac{27}{8}\\\\
x= \sqrt[3]{ \dfrac{27}{8} }\\\\
x= \dfrac{3}{2}

>>>>>>>>>>>>>>>>>>>>>>>>>

log_3x=4\\
x=3^4\\
x=81

>>>>>>>>>>>>>>>>>>>>>>>>>

log_2 \dfrac{1}{16}=x\\\\
2^x= \dfrac{1}{16}\\\\
\not2^x=\not2^{-4}\\\\
x=-4

>>>>>>>>>>>>>>>>>>>>>>>>>

log_x \dfrac{27}{8}=3\\\\
x^3= \dfrac{27}{8}\\\\
x= \sqrt[3]{ \dfrac{27}{8} }\\\\
x= \dfrac{3}{2}

>>>>>>>>>>>>>>>>>>>>>>>>>

log_{ \tfrac{2}{5}}x=-1\\\\
x= \dfrac{2}{5}^{-1}\\\\
x= \dfrac{5}{2}^1\\\\
x= \dfrac{5}{2}

Tenha ótimos estudos ;D
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