Question

log de ⁴√125/8 na base de 4/25. Por favor!

Answer 1

→Para resolver essa questão vou utilizar basicamente as propriedades de potência em logaritmos

$Log_{ \frac{4}{25} } \sqrt[4]{ \frac{125}{8} }$
$Log_{ (\frac{2}{5} )^2} [ (\frac{5}{2} )^3]^\frac{1}{4}$
$\frac{1}{2} Log_{ \frac{2}{5} } ( \frac{5}{2} )^\frac{3}{4}$
$\frac{1}{2} .\frac{3}{4} Log_{ \frac{2}{5} } \frac{5}{2}$
$\frac{3}{8}.Log_{ \frac{2}{5} } \frac{5}{2}$
$\frac{3}{8} . ( \frac{Log \frac{5}{2} }{Log \frac{2}{5} } )$
$\frac{3}{8} . ( \frac{Log 5 - Log 2}{Log 2 - log 5})$
$\frac{3}{8} . \frac{-1.(Log2 - Log5)}{Log2 - log5}$
$\frac{3}{8} . ( - 1 )$
$- \frac{3}{8}$

Answer 2

$\large\begin{array}{l} \textsf{Calcular }\mathsf{\ell og_{\frac{4}{25}}\;^{4}\!\!\!\!\sqrt{\dfrac{125}{8}}.} \end{array}$

$\large\begin{array}{l} \textsf{Chamemos}\\\\ \mathsf{\ell og_{\frac{4}{25}}\;^4\!\!\!\!\sqrt{\dfrac{125}{8}}=x} \end{array}$


Por definição, devemos ter

$\large\begin{array}{l} \mathsf{\left(\dfrac{4}{25}\right)^{\!\!x}=\;^4\!\!\!\!\sqrt{\dfrac{125}{8}}}\\\\ \mathsf{\left(\dfrac{2^2}{5^2}\right)^{\!\!x}=\;^4\!\!\!\!\sqrt{\dfrac{5^3}{2^3}}} \end{array}$

$\large\begin{array}{l}\mathsf{\bigg[\left(\dfrac{2}{5}\right)^{\!\!2}\bigg]^{\!x}=\;^{4}\!\!\!\!\sqrt{\bigg(\dfrac{5}{2}\bigg)^{\!\!3}}}\\\\ \mathsf{\left(\dfrac{2}{5}\right)^{\!\!2x}=\left(\dfrac{5}{2}\right)^{\!\!\frac{3}{4}}}\\\\ \end{array}$

$\large\begin{array}{l} \mathsf{\left(\dfrac{2}{5}\right)^{\!\!2x}=\bigg(\dfrac{~1~}{\frac{2}{5}}\bigg)^{\!\!\frac{3}{4}}}\\\\ \mathsf{\left(\dfrac{2}{5}\right)^{\!\!2x}=\bigg[\left(\dfrac{2}{5}\right)^{\!\!-1}\bigg]^{\frac{3}{4}}}\\\\ \mathsf{\left(\dfrac{2}{5}\right)^{\!\!2x}=\left(\dfrac{2}{5}\right)^{\!\!-\frac{3}{4}}} \end{array}$


$\large\begin{array}{l} \textsf{Na \'ultima linha acima, temos uma igualdade entre exponenciais}\\\textsf{de mesma base. Como a fun\c{c}\~ao exponencial \'e injetora, \'e s\'o}\\\textsf{igualar os expoentes:}\\\\ \mathsf{2x=-\,\dfrac{3}{4}}\\\\ \mathsf{x=-\,\dfrac{3}{4}\cdot \dfrac{1}{2}}\\\\ \mathsf{x=-\,\dfrac{3}{8}}\\\\\\ \therefore~~\boxed{\begin{array}{c}\mathsf{\ell og_{\frac{4}{25}}\;^{4}\!\!\!\!\sqrt{\dfrac{125}{8}}=-\,\dfrac{3}{8}} \end{array}} \end{array}$


$\large\begin{array}{l} \textsf{D\'uvidas? Comente.}\\\\\\ \textsf{Bons estudos! :-)} \end{array}$