# Ex 13.1 Q5 Exponents and Powers Solution- NCERT Maths Class 7

## Question

Express each of the following as product of powers of their prime factors:

(i) \(648\)

(ii) \(405\)

(iii) \(540 \)

(iv) \(3,600\)

## Text Solution

**What is known?**

Numbers.

**What is unknown?**

Prime factors and powers of given numbers.

**Reasoning:**

In this question, first find prime factors of the given number and then raise the prime factor to its powers.

**Steps:**

(i) \(648\)

Prime factorization of \(648\) \(= 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 3\)

In this, \(3\) is the exponent of \(2\) and \(4\) is the exponent of \(3\)

So, \(648\) can be expressed as product of powers of their prime factors as \(2^3 \times\)\(3^4\)

(ii) \(405\)

Prime factorization of \(405\) \(= 3 \times3 \times 3 \times 3 \times 5\)

In this, \(4\) is the exponent of \(3\) and \(1\) is the exponent of \(5\)

So, product of powers of prime factors \(=\) \(3^3 \times\) \(5\)

(iii) \(540\)

Prime factorization of \(540\) \(= 2 \times 2 \times 3 \times 3 \times 3 \times 5\)

In this, \(2\) is the exponent of \(2\). \(3\) is the exponent of \(3\) and \(1\) is the exponent of \(5\)

So, product of powers of prime factors \(= 2^2 \times 3^3 \times 5\)

(iv) \(3,600\)

Prime factorization of \(540\) \(= \left[ \begin{array}{l}2 \times 2 \times 2 \times 2 \times \\3 \times 3 \times 5 \times 5\end{array} \right]\)

In this, \(4\) is the exponent of \(2\). \(2\) is the exponent of \(3\) and \(2\) is the exponent of \(5\)

So, product of powers of prime factors \(= 2^4 \times 3^2 \times 5^2 \)