Given $4725 = 3^a5^b7^c$, find the integral values of $a, b$ and $c$.
Given:
$4725 = 3^a5^b7^c$
To do:
We have to find the integral values of $a, b$ and $c$.
Solution:
Prime factorisation of 4725 is,
$4725=3^3\times5^2\times7^1$
Therefore,
$3^3\times5^2\times7^1=3^a \times 5^b \times 7^c$
Comparing both the sides, we get,
$a=3, b=2$ and $c=1$.
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