1.2.2 Indices, PT3 Practice - Mathematics

Question 6:

Find the value of the following.

(a)

8^{\frac{4}{3}} \times {(3^{- 2})}^{3} \times 9^{\frac{3}{2}}

$8^{\frac{4}{3}} \times {(3^{- 2})}^{3} \times 9^{\frac{3}{2}}$

(b)

\frac{2^{- 2} \times 3^{2}}{2^{- 3} \times 81}

$\frac{2^{- 2} \times 3^{2}}{2^{- 3} \times 81}$

Solution:

(a)

\begin{array}{l} 8^{\frac{4}{3}} \times {(3^{- 2})}^{3} \times 9^{\frac{3}{2}} \\ = {(2^{3})}^{\frac{4}{3}} \times 3^{- 6} \times {(3^{2})}^{\frac{3}{2}} \\ = 2^{4} \times 3^{- 6} \times 3^{3} \\ = 16 \times 3^{- 6 + 3} \\ = 16 \times 3^{- 3} \\ = 16 \times \frac{1}{3^{3}} \\ = \frac{16}{27} \end{array}

$\begin{array}{l} 8^{\frac{4}{3}} \times {(3^{- 2})}^{3} \times 9^{\frac{3}{2}} \\ = {(2^{3})}^{\frac{4}{3}} \times 3^{- 6} \times {(3^{2})}^{\frac{3}{2}} \\ = 2^{4} \times 3^{- 6} \times 3^{3} \\ = 16 \times 3^{- 6 + 3} \\ = 16 \times 3^{- 3} \\ = 16 \times \frac{1}{3^{3}} \\ = \frac{16}{27} \end{array}$

(b)

\begin{array}{l} \frac{2^{- 2} \times 3^{2}}{2^{- 3} \times 81} = \frac{2^{- 2} \times 3^{2}}{2^{- 3} \times 3^{4}} \\ = 2^{- 2 - (- 3)} \times 3^{2 - 4} \\ = 2 \times 3^{- 2} \\ = \frac{2}{3^{2}} \\ = \frac{2}{9} \end{array}

$\begin{array}{l} \frac{2^{- 2} \times 3^{2}}{2^{- 3} \times 81} = \frac{2^{- 2} \times 3^{2}}{2^{- 3} \times 3^{4}} \\ = 2^{- 2 - (- 3)} \times 3^{2 - 4} \\ = 2 \times 3^{- 2} \\ = \frac{2}{3^{2}} \\ = \frac{2}{9} \end{array}$

Question 7:
Given that

2^{8 - x} =32

$2^{8 - x} =32$ , calculate the value of x.

Solution:

\begin{array}{l} 2^{8 - x} =32 \\ 2^{8 - x} {=2}^{5} \\ 8 - x = 5 \\ - x = - 3 \\ x = 3 \end{array}

$\begin{array}{l} 2^{8 - x} =32 \\ 2^{8 - x} {=2}^{5} \\ 8 - x = 5 \\ - x = - 3 \\ x = 3 \end{array}$

Question 8:

Given 3^{2 p - 1} = (3^{p}) (3^{2}), calculate the value of p .

$Given 3^{2 p - 1} = (3^{p}) (3^{2}), calculate the value of p .$

Solution:

\begin{array}{l} 3^{2 p - 1} = (3^{p}) (3^{2}) \\ 3^{2 p - 1} = 3^{p + 2} \\ 2 p - 1 = p + 2 \\ p = 3 \end{array}

$\begin{array}{l} 3^{2 p - 1} = (3^{p}) (3^{2}) \\ 3^{2 p - 1} = 3^{p + 2} \\ 2 p - 1 = p + 2 \\ p = 3 \end{array}$

Question 9:

Given that 8 \times 8^{p + 1} = (8^{5}) (8^{3}), find the value of p .

$Given that 8 \times 8^{p + 1} = (8^{5}) (8^{3}), find the value of p .$

Solution:

\begin{array}{l} 8 \times 8^{p + 1} = (8^{5}) (8^{3}) \\ 8^{1 + p + 1} = 8^{5 + 3} \\ 2 + p = 8 \\ p = 6 \end{array}

$\begin{array}{l} 8 \times 8^{p + 1} = (8^{5}) (8^{3}) \\ 8^{1 + p + 1} = 8^{5 + 3} \\ 2 + p = 8 \\ p = 6 \end{array}$

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