If $1176 = 2^a \times 3^b \times 7^c$, find $a, b$ and $c$.

Given:

$1176 = 2^a \times 3^b \times 7^c$

To do:

We have to find the values of $a, b$ and $c$.

Solution:

Prime factorisation of 1176 is,

$1176=2^3\times3^1\times7^2$

Therefore,

$2^3\times3^1\times7^2=2^a \times 3^b \times 7^c$

Comparing both the sides, we get,

$a=3, b=1$ and $c=2$. 

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Updated on: 10-Oct-2022

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