Answer
If $f(\alpha{x})=a^{\alpha{x}}=\underbrace{a^x \times a^x \times a^x . . . \times a^x}=\underbrace{f(x)\times f(x) \times f(x) ... \times f(x)}=[f(x)]^{\alpha}\\
\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \alpha \text{ times} \quad \quad \quad \quad \quad \quad \quad \quad \alpha \text{ times}$.
Work Step by Step
Given:
$f(x)=a^{x}$...........$(1)$
Then,
$f(\alpha x)= a^{\alpha x}$...........[from (1)]
$f(\alpha x)= \underbrace{{a^{x}\times a^{x}\times ....\times a^{x}}}\\
\quad \quad \quad \quad \quad \text{$(\alpha$ times})$........[by laws of exponents]
$f(\alpha x)= {f(x)\times f(x)\times ....\times f(x)
} \quad \text{$(\alpha$ times})$ ..........[by using (1)]
$f(\alpha x)=[f(x)]^{\alpha}$.........[by using laws of exponents]
Hence Proved.
