1.2.4 Indices, PT3 Practice - Mathematics

Question 14:

Given that k^{3} = 9^{- \frac{3}{2}} \times 64^{\frac{1}{2}}, find the value of k .

$Given that k^{3} = 9^{- \frac{3}{2}} \times 64^{\frac{1}{2}}, find the value of k .$

Solution:

\begin{matrix} k^{3} = 9^{- \frac{3}{2}} \times 64^{\frac{1}{2}} = {(3^{2})}^{- \frac{3}{2}} \times {(2^{6})}^{\frac{1}{2}} = 3^{- 3} \times 2^{3} = {(\frac{2}{3})}^{3} k = \frac{2}{3} \end{matrix}

$\begin{array}{l} k^{3} = 9^{- \frac{3}{2}} \times 64^{\frac{1}{2}} \\ = {(3^{2})}^{- \frac{3}{2}} \times {(2^{6})}^{\frac{1}{2}} \\ = 3^{- 3} \times 2^{3} \\ = {(\frac{2}{3})}^{3} \\ k = \frac{2}{3} \end{array}$

Question 15:

{Given 9}^{x + 2} \div 3^{4} = 3^{x + 1}, calculate the value of x .

${Given 9}^{x + 2} \div 3^{4} = 3^{x + 1}, calculate the value of x .$

Solution:

\begin{matrix} 9^{x + 2} \div 3^{4} = 3^{x + 1} {(3^{2})}^{x + 2} \div 3^{4} = 3^{x + 1} 3^{2 x + 4} \div 3^{4} = 3^{x + 1} 2 x + 4 - 4 = x + 1 2 x = x + 1 x = 1 \end{matrix}

$\begin{array}{l} 9^{x + 2} \div 3^{4} = 3^{x + 1} \\ {(3^{2})}^{x + 2} \div 3^{4} = 3^{x + 1} \\ 3^{2 x + 4} \div 3^{4} = 3^{x + 1} \\ 2 x + 4 - 4 = x + 1 \\ 2 x = x + 1 \\ x = 1 \end{array}$

Question 16:

Simplify: {(\frac{- 2 x^{5} y^{- 2}}{z^{\frac{1}{6}}})}^{3} \div \frac{1}{\sqrt{x^{2} z}}

$Simplify: {(\frac{- 2 x^{5} y^{- 2}}{z^{\frac{1}{6}}})}^{3} \div \frac{1}{\sqrt{x^{2} z}}$

Solution:

$\begin{array}{l} {(\frac{- 2 x^{5} y^{- 2}}{z^{\frac{1}{6}}})}^{3} \div \frac{1}{\sqrt{x^{2} z}} \\ = \frac{- 8 x^{15} y^{- 6}}{z^{\frac{1}{2}}} \times \sqrt{x^{2} z} \\ = \frac{- 8 x^{15} y^{- 6}}{z^{\frac{1}{2}}} \times {(x^{2} z)}^{\frac{1}{2}} \\ = \frac{- 8 x^{15} y^{- 6}}{z^{\frac{1}{2}}} \times x z^{\frac{1}{2}} \\ = - 8 x^{15 + 1} y^{- 6} \\ = \frac{- 8 x^{16}}{y^{6}} \end{array}$

Question 17:

Find the value of the following.

(a) (2³)²× 2⁴÷ 2⁵

(b)

$\frac{a^{2} \times a^{\frac{1}{2}}}{{(a^{\frac{2}{3}} \times a^{\frac{1}{3}})}^{- 2}}$

Solution:

(a)

(2³)²× 2⁴÷ 2⁵= 2⁶× 2⁴÷ 2⁵

= 2^6+4-5

= 2⁵

= 32

(b)

$\begin{array}{l} \frac{a^{2} \times a^{\frac{1}{2}}}{{(a^{\frac{2}{3}} \times a^{\frac{1}{3}})}^{- 2}} = \frac{a^{2 + \frac{1}{2}}}{{(a^{\frac{2}{3}} \times a^{\frac{1}{3}})}^{- 2}} \\ = \frac{a^{2 + \frac{1}{2}}}{{(a^{\frac{2}{3} + \frac{1}{3}})}^{- 2}} \\ = \frac{a^{\frac{5}{2}}}{a^{- 2}} \\ = a^{\frac{5}{2} - (- 2)} \\ = a^{\frac{5}{2} + \frac{4}{2}} \\ = a^{\frac{9}{2}} \end{array}$

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