We know in general,

\(a^n=\underbrace{a\times a\times a\times ... \times a}\space; a\neq0. \\ \qquad \qquad \space n \text{ factor}\)

Note that in index notation, the number of index is equal to the number of bases multiplied by itself repeatedly.

Therefore,

\(\begin{aligned} \left(\dfrac{4}{5}\right)^3&=\underbrace{\dfrac{4}{5}\times \dfrac{4}{5}}\times \space\dfrac{4}{5} \\\\&= \space \space \space \dfrac{16}{25} \space \space\times\dfrac{4}{5}\\\\ &=\dfrac{64}{125}. \end{aligned}\)

Kita mengetahui secara generalisasi,

\(a^n=\underbrace{a\times a\times a\times ... \times a}\space; a\neq0 \\\qquad \qquad \space n \text{ faktor}\)

Perhatikan bahawa dalam tatatanda indeks, bilangan indeks adalah sama dengan bilangan asas didarab secara berulang.

Maka,

\(\begin{aligned} \left(\dfrac{4}{5}\right)^3&=\underbrace{\dfrac{4}{5}\times \dfrac{4}{5}}\times \space\dfrac{4}{5} \\\\&= \space \space \space \dfrac{16}{25} \space \space\times\dfrac{4}{5}\\\\ &=\dfrac{64}{125}. \end{aligned}\)

We know in general,

\(a^n=\underbrace{a\times a\times a\times ... \times a}\space; a\neq0. \\ \qquad \qquad \space n \text{ factor}\)

Note that in index notation, the number of index is equal to the number of bases multiplied by itself repeatedly.

Therefore,

\(\begin{aligned} \left(\dfrac{4}{5}\right)^3&=\underbrace{\dfrac{4}{5}\times \dfrac{4}{5}}\times \space\dfrac{4}{5} \\\\&= \space \space \space \dfrac{16}{25} \space \space\times\dfrac{4}{5}\\\\ &=\dfrac{64}{125}. \end{aligned}\)