Given 4725=3a5b7c, find


Given $4725=3^{a} 5^{b} 7^{c}$, find

(i) the integral values of $a, b$ and $c$

(ii) the value of $2^{-a} 3^{b} 7^{c}$



(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 $c=1$


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

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

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

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


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

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