1.2.3 Indices, PT3 Practice - Mathematics

Question 10:

Given that \frac{2^{5} \times 2^{7}}{2^{10}} = 2^{p}, find the value of p .

$Given that \frac{2^{5} \times 2^{7}}{2^{10}} = 2^{p}, find the value of p .$

Solution:

\begin{matrix} \frac{2^{5} \times 2^{7}}{2^{10}} = 2^{p} 2^{5 + 7 - 10} = 2^{p} 2^{2} = 2^{p} p = 2 \end{matrix}

$\begin{array}{l} \frac{2^{5} \times 2^{7}}{2^{10}} = 2^{p} \\ 2^{5 + 7 - 10} = 2^{p} \\ 2^{2} = 2^{p} \\ p = 2 \end{array}$

Question 11:

Simplify: \frac{12 p^{10} q^{6}}{3 p^{4} q^{2} \times {(4 p q^{3})}^{2}}

$Simplify: \frac{12 p^{10} q^{6}}{3 p^{4} q^{2} \times {(4 p q^{3})}^{2}}$

Solution:

\begin{matrix} \frac{12 p^{10} q^{6}}{3 p^{4} q^{2} \times {(4 p q^{3})}^{2}} = \frac{12 p^{10} q^{6}}{3 p^{4} q^{2} \times 16 p^{2} q^{6}} = \frac{12^{1} p^{10 - 4 - 2} q^{6 - 2 - 6}}{3^{1} \times 16^{4}} = \frac{p^{4} q^{- 2}}{4} = \frac{p^{4}}{4 q^{2}} \end{matrix}

$\begin{array}{l} \frac{12 p^{10} q^{6}}{3 p^{4} q^{2} \times {(4 p q^{3})}^{2}} = \frac{12 p^{10} q^{6}}{3 p^{4} q^{2} \times 16 p^{2} q^{6}} \\ = \frac{12^{1} p^{10 - 4 - 2} q^{6 - 2 - 6}}{3^{1} \times 16^{4}} \\ = \frac{p^{4} q^{- 2}}{4} \\ = \frac{p^{4}}{4 q^{2}} \end{array}$

Question 12:

Simplify: \frac{a b^{2} \times {(8 a^{3} b^{6})}^{\frac{1}{3}}}{{(a^{2} b^{4})}^{\frac{1}{2}}}

$Simplify: \frac{a b^{2} \times {(8 a^{3} b^{6})}^{\frac{1}{3}}}{{(a^{2} b^{4})}^{\frac{1}{2}}}$

Solution:

\begin{matrix} \frac{a b^{2} \times {(8 a^{3} b^{6})}^{\frac{1}{3}}}{{(a^{2} b^{4})}^{\frac{1}{2}}} = \frac{a b^{2} \times (8^{^{\frac{1}{3}}} a^{3}^{(\frac{1}{3})} b^{6 (\frac{1}{3})})}{a^{2 (\frac{1}{2})} b^{4 (\frac{1}{2})}} = \frac{a b^{2} \times 2 a b^{2}}{a b^{2}} = 2 a b^{2} \end{matrix}

$\begin{array}{l} \frac{a b^{2} \times {(8 a^{3} b^{6})}^{\frac{1}{3}}}{{(a^{2} b^{4})}^{\frac{1}{2}}} = \frac{a b^{2} \times (8^{^{\frac{1}{3}}} a^{3}^{(\frac{1}{3})} b^{6 (\frac{1}{3})})}{a^{2 (\frac{1}{2})} b^{4 (\frac{1}{2})}} \\ = \frac{a b^{2} \times 2 a b^{2}}{a b^{2}} \\ = 2 a b^{2} \end{array}$

Question 13:

Solve: 3^{2 n + 1} = \frac{3^{n} \times 9}{{(9^{\frac{1}{2}})}^{- 3}}

$Solve: 3^{2 n + 1} = \frac{3^{n} \times 9}{{(9^{\frac{1}{2}})}^{- 3}}$

Solution:

\begin{matrix} 3^{2 n + 1} = \frac{3^{n} \times 9}{{(9^{\frac{1}{2}})}^{- 3}} 3^{2 n + 1} = \frac{3^{n} \times 3^{2}}{3^{- 3}} 3^{2 n + 1} = 3^{n + 2 - (- 3)} 2 n + 1 = n + 5 n = 4 \end{matrix}

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

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