Question

Gente, fiz essa equação até aqui, está correto? Como continuar? Se estiver errada, favor, corrigir.
(Função exponencial)

$\sqrt[3]{4 ^{x} } = \frac{1}{8} \sqrt[3]{2 ^{2x} } = \frac{1}{2^3} \sqrt[3]{2 ^{2x} } = \frac{2}{2^-^3} 2^{-2x\frac{-2}{3} = 2^{-3}$

Answer 1

$\textsf{Principais propriedades exponenciais:}\\\\\\\mathsf{P1.~~a^m\cdot a^n=a^{m+n}}\\\\\\\mathsf{P2.~~\dfrac{a^m}{a^n}=a^{m-n}~~~~(para~a \neq0~e~m \geq n)}\\\\\\\mathsf{P3.~~(a\cdot b)^n=a^n\cdot b^n}\\\\\\\mathsf{P4.~~(\dfrac{a}{b})^n=\dfrac{a^n}{b^n}~~~~(b \neq 0)}\\\\\\\mathsf{P5.~~(a^m)^n=a^{m\cdot n}}\\\\\\\textsf{Expoente racional:}~~\fbox{ \mathsf{a^{\frac{b}{c}}=\sqrt[c]{\mathsf{a^b}}} }\\\\\\\textsf{Expoente inteiro negativo:}~~\fbox{ \mathsf{a^{-n}=\dfrac{1}{a^n}} }$

$\begin{array}{l}\textsf{Para essa quest\~ao vamos utilizar das propriedades P5, expoente racional e}\\\textsf{expoente inteiro negativo. Teremos o seguinte:}\end{array}$

$\large\begin{array}{l}\mathsf{\sqrt[\mathsf{3}]{\mathsf{4^x}}=\dfrac{1}{8} ~\Leftrightarrow~4^{\frac{x}{3}}=8^{-1}~\Leftrightarrow~(2^2)^{\frac{x}{3} }=(2^3)^{-1}}\\\\\mathsf{2^{2\cdot\frac{x}{3}}=2^{3\cdot(-1)}~\Leftrightarrow~2^{ \frac{2x}{3}}=2^{-3}}\end{array}$

$\begin{array}{l}\mathsf{Tendo~em~mente~que~em~uma~equa\c{c}\~ao~exponencial}\\\\\fbox{ \mathsf{a^b=a^c~\Longleftrightarrow~b=c~~~~(0\ \textless \ a \neq 1)} }\\\\\mathsf{ent\~ao:}\end{array}$

$\large\begin{array}{l}\mathsf{2^{ \frac{2x}{3} }=2^{-3}~\Leftrightarrow~ \dfrac{2x}{3}=-3~\Leftrightarrow~2x=-9~\Leftrightarrow~\fbox{ \mathsf{x= -\dfrac{9}{2}} }}\end{array}$