Question

Given that
a = 1/64x^2 b = 16/√x
express each of the following in the form kx^n where k and n are simplified constants
a) a^1/2
b) 16/b^3
c) (ab/2)^-4/3

Answer 1

Answer:

$\textsf{(a)}\quad \dfrac{1}{8}x$

$\textsf{(b)}\quad \dfrac{1}{256}x^{\frac32}$

$\textsf{(c)}\quad 16x^{-2}$

Step-by-step explanation:

To rewrite the given expressions in the form of $kx^n$, where k and n are simplified constants, we can use the rules of exponents.

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Part (a)

$\textsf{Given that\;\; a=\dfrac{1}{64}x^2 , then:}$

$a^{\frac{1}{2}}=\left(\dfrac{1}{64}x^2\right)^{\frac{1}{2}}$

$\textsf{Apply the power of a product exponent rule:} \quad (ab)^c=a^{c}b^c$

      $=\left(\dfrac{1}{64}\right)^{\frac{1}{2}}(x^2)^{\frac{1}{2}}$

$\textsf{Apply the power of a quotient exponent rule:} \quad \left(\dfrac{a}{b}\right)^c=\dfrac{a^c}{b^c}$

      $=\left(\dfrac{1^{\frac{1}{2}}}{64^{\frac{1}{2}}}\right)(x^2)^{\frac{1}{2}}\\\\\\=\dfrac{1}{8}(x^2)^{\frac12}$

$\textsf{Apply the power of a power exponent rule:} \quad (a^b)^c=a^{bc}$

      $=\dfrac{1}{8}x^{(2 \times \frac12)}\\\\\\=\dfrac{1}{8}x^{1}\\\\\\=\dfrac{1}{8}x$

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Part (b)

$\textsf{Given that\;\; b=\dfrac{16}{\sqrt{x}} ,\;then:}$

$\dfrac{16}{b^3}=\dfrac{16}{\left(\dfrac{16}{\sqrt{x}}\right)^3}$

$\textsf{Rewrite \sqrt{x} as x^{\frac{1}{2}} :}$

    $=\dfrac{16}{\left(\dfrac{16}{x^{\frac12}}\right)^3}$

$\textsf{Apply the power of a quotient exponent rule:} \quad \left(\dfrac{a}{b}\right)^c=\dfrac{a^c}{b^c}$

     $=\dfrac{16}{\left(\dfrac{16^3}{\left(x^{\frac12}}\right)^3\right)}\\\\\\=\dfrac{16}{\left(\dfrac{4096}{\left(x^{\frac12}}\right)^3\right)}$

$\textsf{Apply the power of a power exponent rule:} \quad (a^b)^c=a^{bc}$

     $=\dfrac{16}{\left(\dfrac{4096}{x^{\frac32}}\right)}$

$\textsf{Apply the fraction rule:} \quad \dfrac{a}{\frac{b}{c}}=\dfrac{ac}{b}$

     $=\dfrac{16x^{\frac32}}{4096}\\\\\\=\dfrac{1}{256}x^{\frac32}$

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Part (c)

$\textsf{Given that\;\; a=\dfrac{1}{64}x^2 \;\;and\;\; b=\dfrac{16}{\sqrt{x}} ,\;then:}$

$\left(\dfrac{ab}{2}\right)^{-\frac{4}{3}}=\left(\dfrac{\left(\dfrac{1}{64}x^2\right)\left(\dfrac{16}{\sqrt{x}}\right)}{2}\right)^{-\frac{4}{3}}$

Simplify the numerator:

               $=\left(\dfrac{\left(\dfrac{16x^2}{64\sqrt{x}}\right)}{2}\right)^{-\frac{4}{3}}\\\\\\\\=\left(\dfrac{\left(\dfrac{x^{\left(2-\frac12\right)}}{4}\right)}{2}\right)^{-\frac{4}{3}}\\\\\\\\=\left(\dfrac{\left(\dfrac{x^{\frac32}}{4}\right)}{2}\right)^{-\frac{4}{3}}$

$\textsf{Apply the fraction rule:} \quad \dfrac{\frac{a}{b}}{c}=\dfrac{a}{bc}$

               $=\left(\dfrac{x^{\frac32}}{4 (2)}\right)^{-\frac{4}{3}}\\\\\\\\=\left(\dfrac{x^{\frac32}}{8}\right)^{-\frac{4}{3}}$

$\textsf{Apply the power of a quotient exponent rule:} \quad \left(\dfrac{a}{b}\right)^c=\dfrac{a^c}{b^c}$

               $=\dfrac{\left(x^{\frac32}\right)^{-\frac43}}{(8)^{-\frac43}}\\\\\\\\=\dfrac{\left(x^{\frac32}\right)^{-\frac43}}{\frac{1}{16}}$

$\textsf{Apply the power of a power exponent rule:} \quad (a^b)^c=a^{bc}$

               $=\dfrac{x^{\left(\frac{3}{2}\left(-\frac{4}{3}\right)}\right)}{\frac{1}{16}}\\\\\\\\=\dfrac{x^{-\frac{12}{6}}}{\frac{1}{16}}\\\\\\\\=\dfrac{x^{-2}}{\frac{1}{16}}$

$\textsf{Apply the fraction rule:} \quad \dfrac{a}{\frac{b}{c}}=\dfrac{ac}{b}$

               $=\dfrac{16x^{-2}}{1}\\\\\\=16x^{-2}$