Question:
Given $4725=3^{a} 5^{b} 7^{c}$, find
(i) the integral values of $a, b$ and $c$
(ii) the value of $2^{-a} 3^{b} 7^{c}$
Solution:
(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 $\frac{63}{8}$.