1.2.1 Indices, PT3 Practice

June 27, 2020May 15, 2017 by

1.2.1 Indices, PT3 Practice

Question 1:

(a) Simplify: a⁴÷ a⁷

(b) Evaluate:

{(2^{4})}^{\frac{1}{2}} \times 3^{\frac{1}{2}} \times 12^{\frac{1}{2}}

Solution:

(a) a⁴÷ a⁷= a^4-7= a^-3

(b)

\begin{array}{l} {(2^{4})}^{\frac{1}{2}} \times 3^{\frac{1}{2}} \times 12^{\frac{1}{2}} = 2^{2} \times 3^{\frac{1}{2}} \times {(4 \times 3)}^{\frac{1}{2}} \\ = 2^{2} \times 3^{\frac{1}{2}} \times {(2^{2} \times 3)}^{\frac{1}{2}} \\ = 2^{2} \times 3^{\frac{1}{2}} \times 2 \times 3^{\frac{1}{2}} \\ = 2^{3} \times 3 \\ = 24 \end{array}

Question 2:

(a) Simplify: p³÷ p^-5

(b) Evaluate:

10^{\frac{1}{2}} \times 5^{- \frac{1}{2}} \times {(2^{\frac{1}{2}})}^{5}

Solution:

(a) p³÷ p^-5= p^3-(-5)= p³⁺⁵= p⁸

(b)

\begin{array}{l} 10^{\frac{1}{2}} \times 5^{- \frac{1}{2}} \times {(2^{\frac{1}{2}})}^{5} \\ = {(2 \times 5)}^{\frac{1}{2}} \times 5^{- \frac{1}{2}} \times 2^{\frac{5}{2}} \\ = 2^{\frac{1}{2}} \times 5^{\frac{1}{2}} \times 5^{- \frac{1}{2}} \times 2^{\frac{5}{2}} \\ = 2^{\frac{1}{2} + \frac{5}{2}} \times 5^{\frac{1}{2} + (- \frac{1}{2})} \\ = 2^{3} + 5^{\frac{1}{2} - \frac{1}{2}} \\ = 2^{3} + 5^{0} \\ = 8 + 1 \\ = 9 \end{array}

Question 3:

(a) Find the value of

10^{\frac{4}{3}} \div 10^{\frac{1}{3}} .

(b) Simplify (xy³)⁵ × x⁴.

Solution:

(a)

\begin{array}{l} 10^{\frac{4}{3}} \div 10^{\frac{1}{3}} \\ = 10^{\frac{4}{3} - \frac{1}{3}} \\ = 10^{\frac{3}{3}} \\ = 10 \end{array}

\begin{array}{l} (b) {(x y^{3})}^{5} \times x^{4} = x^{5} y^{15} \times x^{4} \\ = x^{5 + 4} y^{15} \\ = x^{9} y^{15} \end{array}

Question 4:

(a)

{(81 a^{8})}^{- \frac{1}{4}} =

(b) Find the value of 2³× 2²

Solution:

(a)

{(81 a^{8})}^{- \frac{1}{4}} = \frac{1}{{(81 a^{8})}^{\frac{1}{4}}} = \frac{1}{{(3^{4})}^{\frac{1}{4}} {(a^{8})}^{\frac{1}{4}}} = \frac{1}{3 a^{2}}

(b)

2³× 2²= 2³⁺²= 2⁵= 32

Question 5:

Find the value of the following.

(a)

81^{\frac{3}{4}} \times 27^{- 1}

(b)

8^{\frac{2}{3}} \times 3^{- 2}

Solution:

(a)

\begin{array}{l} 81^{\frac{3}{4}} \times 27^{- 1} = {(\sqrt[4]{81})}^{3} \times {(3^{3})}^{- 1} \\ = {(3)}^{3} \times 3^{- 3} \\ = 3^{3 + (- 3)} \\ = 3^{0} = 1 \end{array}

(b)

\begin{array}{l} 8^{\frac{2}{3}} \times 3^{- 2} = {(\sqrt[3]{8})}^{2} \times \frac{1}{3^{2}} \\ = {(2)}^{2} \times \frac{1}{3^{2}} \\ = 4 \times \frac{1}{9} \\ = \frac{4}{9} \end{array}

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