1.2.2 Indeks, Praktis Berformat PT3

June 21, 2020September 17, 2017 by

Soalan 5:

Cari nilai bagi setiap yang berikut.

(a)

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

(b)

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

Penyelesaian:

(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}

Soalan 6:

Cari nilai bagi setiap yang berikut.

(a)

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

(b)

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

Penyelesaian:

(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}

(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}

Soalan 7:
Diberi bahawa

2^{8 - x} =32

, hitung nilai bagi x.

Penyelesaian:

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

Soalan 8:

Diberi 3^{2 p - 1} = (3^{p}) (3^{2}), hitung nilai bagi p .

Penyelesaian:

\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}

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