If $1176 = 2^a \times 3^b \times 7^c$, find the values of $a, b$ and $c$. Hence, compute the value of $2^a \times 3^b \times 7^{-c}$ as a fraction.

Academic Mathematics NCERT Class 9

Given:

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

To do:

We have to find $a, b$ and $c$ and compute the value of $2^a \times 3^{b} \times 7^{-c}$.

Solution:

We know that,

$(a^{m})^{n}=a^{m n}$

$a^{m} \times a^{n}=a^{m+n}$

$a^{m} \div a^{n}=a^{m-n}$

$a^{0}=1$

Therefore,

Prime factorisation of 1176 is,

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

This implies,

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

Comparing both sides, we get,

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

This implies,

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

$=\frac{2^3\times3^1}{7^2}$

$=\frac{8\times3}{49}$

$=\frac{24}{49}$

The values of $a, b$ and $c$ are 3, 1 and 2 respectively. The value of $2^a \times 3^{b} \times 7^{-c}$ is $\frac{24}{49}$.

Simply Easy Learning

Updated on: 10-Oct-2022

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