Given 4725=3a5b7c, find

(i) Given $4725=3^{a} 5^{b} 7^{c}$

First find out the prime factorisation of 4725.

It can be observed that 4725 can be written as $3^{3} \times 5^{2} \times 7^{1}$

$\therefore 4725=3^{a} 5^{b} 7^{c}=3^{3} 5^{2} 7^{1}$

Hence, $a=3, b=2$ and $c=1$

(ii)

When $a=3, b=2$ and $c=1$,

$2^{-a} 3^{b} 7^{c}$

$=2^{-3} \times 3^{2} \times 7^{1}$

$=\frac{1}{8} \times 9 \times 7$

$=\frac{63}{8}$

Hence, the value of $2^{-a} 3^{b} 7^{c}$ is $\frac{63}{8}$.