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If $x$ is numerically so small so that $x^2$ and higher powers of $x$ can be neglected, then $\left(1+\frac{2 x}{3}\right)^{3 / 2} \cdot(32+5 x)^{-1 / 5}$ is approximately equal to
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Verified Answer
The correct answer is:
$\frac{32+31 x}{64}$
$\begin{aligned}
(1+ & \left.\frac{2 x}{3}\right)^{3 / 2}(32+5 x)^{-1 / 5} \\
& =\left[1+\frac{3}{2}\left(\frac{2 x}{3}\right)\right](32)^{-1 / 5}\left(1+\frac{5}{32} x\right)^{-1 / 5}
\end{aligned}$
(Neglect higher powers of $x$ )
$=[1+x] 2^{-1}\left[1-\frac{1}{5}\left(\frac{5}{32}\right) x\right]$
(Neglect higher powers of $x$ )
$\begin{aligned}
& =\frac{1}{2}(1+x)\left(1-\frac{x}{32}\right) \\
& =\frac{(1+x)(32-x)}{64}=\frac{32+31 x}{64}
\end{aligned}$
(Neglect $x^2$ term)
(1+ & \left.\frac{2 x}{3}\right)^{3 / 2}(32+5 x)^{-1 / 5} \\
& =\left[1+\frac{3}{2}\left(\frac{2 x}{3}\right)\right](32)^{-1 / 5}\left(1+\frac{5}{32} x\right)^{-1 / 5}
\end{aligned}$
(Neglect higher powers of $x$ )
$=[1+x] 2^{-1}\left[1-\frac{1}{5}\left(\frac{5}{32}\right) x\right]$
(Neglect higher powers of $x$ )
$\begin{aligned}
& =\frac{1}{2}(1+x)\left(1-\frac{x}{32}\right) \\
& =\frac{(1+x)(32-x)}{64}=\frac{32+31 x}{64}
\end{aligned}$
(Neglect $x^2$ term)
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