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1.4: Exponents
https://math.libretexts.org/Courses/Chabot_College/MTH_15%3A_Applied_Calculus_I/01%3A_Prerequisite_Review/1.04%3A_Exponents
$x^a\cdot x^b=x^{a+b}$ Since $x^{-n}=\dfrac{1}{x^n}$ , then $x^{-3}=\dfrac{1}{x^3}$ and thus $\dfrac{5}{x^3}=5x^{-3}.\nonumber$ \[\begin{align*} x^{-4/3} & = \frac{1}{x^{4/3}} \\ & = \frac{1}{... $x^a\cdot x^b=x^{a+b}$ Since $x^{-n}=\dfrac{1}{x^n}$ , then $x^{-3}=\dfrac{1}{x^3}$ and thus $\dfrac{5}{x^3}=5x^{-3}.\nonumber$ $\begin{align*} x^{-4/3} & = \frac{1}{x^{4/3}} \\ & = \frac{1}{\left(x^{1/3}\right)^4} \quad \text{(since $\frac{4}{3}=4\cdot\frac{1}{3}$)}\\ & = \frac{1}{\left(\sqrt[3]{x}\right)^4} \quad \text{(using the radical equivalence)} \end{align*}$
1.4: Exponents
https://math.libretexts.org/Courses/Penn_State_University_Greater_Allegheny/MATH_110%3A_Techniques_of_Calculus_I_(Gaydos)/01%3A_Review/1.04%3A_Exponents
Since $x^{-n}=\dfrac{1}{x^n}$ , then $x^{-3}=\dfrac{1}{x^3}$ and thus $\dfrac{5}{x^3}=5x^{-3}.\nonumber$ \begin{align*} x^{-4/3} & = \frac{1}{x^{4/3}} \\ & = \frac{1}{\left(x^{1/3}\right)^4} \qu...Since $x^{-n}=\dfrac{1}{x^n}$ , then $x^{-3}=\dfrac{1}{x^3}$ and thus $\dfrac{5}{x^3}=5x^{-3}.\nonumber$ $\begin{align*} x^{-4/3} & = \frac{1}{x^{4/3}} \\ & = \frac{1}{\left(x^{1/3}\right)^4} \quad \text{(since $\frac{4}{3}=4\cdot\frac{1}{3}$)}\\ & = \frac{1}{\left(\sqrt[3]{x}\right)^4} \quad \text{(using the radical equivalence)} \end{align*}$
2.1: Exponents
https://math.libretexts.org/Courses/Prince_Georges_Community_College/MAT_2160%3A_Applied_Calculus_I/02%3A_Exponential_and_Logarithmic_functions/2.01%3A_Exponents
Since $x^{-n}=\dfrac{1}{x^n}$ , then $x^{-3}=\dfrac{1}{x^3}$ and thus $\dfrac{5}{x^3}=5x^{-3}.\nonumber$ \begin{align*} x^{-4/3} & = \frac{1}{x^{4/3}} \\ & = \frac{1}{\left(x^{1/3}\right)^4} \qu...Since $x^{-n}=\dfrac{1}{x^n}$ , then $x^{-3}=\dfrac{1}{x^3}$ and thus $\dfrac{5}{x^3}=5x^{-3}.\nonumber$ $\begin{align*} x^{-4/3} & = \frac{1}{x^{4/3}} \\ & = \frac{1}{\left(x^{1/3}\right)^4} \quad \text{(since $\frac{4}{3}=4\cdot\frac{1}{3}$)}\\ & = \frac{1}{\left(\sqrt[3]{x}\right)^4} \quad \text{(using the radical equivalence)} \end{align*}$
1.4: Exponents
https://math.libretexts.org/Courses/Butler_Community_College/MA148%3A_Calculus_with_Applications_-_Butler_CC/01%3A_Review/1.04%3A_Exponents
$x^a\cdot x^b=x^{a+b}$ Since $x^{-n}=\dfrac{1}{x^n}$ , then $x^{-3}=\dfrac{1}{x^3}$ and thus $\dfrac{5}{x^3}=5x^{-3}.\nonumber$ \[\begin{align*} x^{-4/3} & = \frac{1}{x^{4/3}} \\ & = \frac{1}{... $x^a\cdot x^b=x^{a+b}$ Since $x^{-n}=\dfrac{1}{x^n}$ , then $x^{-3}=\dfrac{1}{x^3}$ and thus $\dfrac{5}{x^3}=5x^{-3}.\nonumber$ $\begin{align*} x^{-4/3} & = \frac{1}{x^{4/3}} \\ & = \frac{1}{\left(x^{1/3}\right)^4} \quad \text{(since $\frac{4}{3}=4\cdot\frac{1}{3}$)}\\ & = \frac{1}{\left(\sqrt[3]{x}\right)^4} \quad \text{(using the radical equivalence)} \end{align*}$

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