Question

Me ajudem por favor; $log_{ \frac{2}{3} } \frac{8}{27}$

Answer 1

$\boxed{log_{b}(a) =x \to b^x=a}\to \left\{\begin{array}{ccc}log_{ \frac{2}{3}} \frac{8}{27}=x \\\\( \frac{2}{3})^x = \frac{8}{27}\\\\ (\frac{2}{3})^x = \frac{2^3}{3^3}\\\\( \frac{2}{3} )^x = ( \frac{2}{3})^3\\\\\boxed{x = 3} \end{array}\right$

Espero ter ajudado. :))

Answer 2

Olá Observadora,

aplique as propriedades, da potência e a decorrente da definição:

$logb^k~\to~k*logb\\\\ log_kk=1$

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$log_{ \tfrac{2}{3} } \dfrac{8}{27}=x\\\\ x=log_{ \tfrac{2}{3} } (\dfrac{2^3}{3^3})\\\\ x= log_{ \tfrac{2}{3} }( \dfrac{2}{3})^3\\\\ x=3*log_{ \tfrac{2}{3} } \dfrac{2}{3}\\\\ x=3*1\\\\ \boxed{x=3}$

Espero ter ajudado e tenha ótimos estudos =))