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Surds and Indices Practice Test 1

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Question : 5 of 40

Marks: +1, -0

(16)^{-\frac{3}{4}} + 2^{-3} + (8)^{-\frac{2}{3}}

is equivalent to:

$\frac{1}{2}$
$\frac{3}{2}$
$\frac{9}{2}$
$\frac{5}{4}$

Solution:

Concept:

Apply the laws of negative and fractional exponents.

Explanation:

Convert each negative power into its reciprocal.

(16)^{-\frac{3}{4}}=\frac{1}{16^{\frac{3}{4}}}=\frac{1}{(2^4)^{\frac{3}{4}}}=\frac{1}{2^3}=\frac{1}{8}

2^{-3}=\frac{1}{2^3}=\frac{1}{8}

(8)^{-\frac{2}{3}}=\frac{1}{8^{\frac{2}{3}}}=\frac{1}{(2^3)^{\frac{2}{3}}}=\frac{1}{2^2}=\frac{1}{4}

Now add all three terms:

\frac{1}{8}+\frac{1}{8}+\frac{1}{4}=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}

Answer:

\frac{1}{2}

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