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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Updated on: 10-Oct-2022

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