Verified Answer :

  • Find the prime factorization of 48 by repeated division: \(48 = 2 \times 2 \times 2 \times 2 \times 3 = 2^4 \times 3\).
  • Simplify the expression \(\frac{5^5}{5^3}\) using the law of exponents \(\frac{a^m}{a^n} = a^{m-n}\): \(5^{5-3} = 5^2 = 25\).
  • Final answers: \(48 = \boxed{2^4 \times 3}\) and \(\frac{5^5}{5^3} = \boxed{25}\).

Explanation

  1. Problem Analysis

We are tasked with two mathematical operations: simplifying the number 48 (which implies finding its prime factorization) and calculating the expression \(\frac{5^5}{5^3}\).
To find the prime factorization of 48, we express it as a product of prime numbers.
For the division of powers, we will utilize the rule of exponents: \(\frac{a^m}{a^n} = a^{m-n}\) for any non-zero base \(a\).

  1. Prime Factorization of 48

To find the prime factorization of 48, we divide by the smallest prime numbers repeatedly:

  1. Divide 48 by 2: \(48 \div 2 = 24\)
  2. Divide 24 by 2: \(24 \div 2 = 12\)
  3. Divide 12 by 2: \(12 \div 2 = 6\)
  4. Divide 6 by 2: \(6 \div 2 = 3\)
  5. Divide 3 by 3: \(3 \div 3 = 1\)

Thus, \(48 = 2 \times 2 \times 2 \times 2 \times 3 = 2^4 \times 3\).

  1. Calculating \(\frac{5^5}{5^3}\)

We need to compute \(\frac{5^5}{5^3}\). Using the quotient rule for exponents, \(\frac{a^m}{a^n} = a^{m-n}\):
\[\frac{5^5}{5^3} = 5^{5-3} = 5^2\]
Evaluating \(5^2\), we get \(5 \times 5 = 25\).

  1. Final Conclusion

The prime factorization of 48 is \(2^4 \times 3\), and the value of \(\frac{5^5}{5^3}\) is 25. Therefore, the simplified form of 48 is \(\boxed{2^4 \times 3}\) and the result of the division is \(\boxed{25}\).

Examples

Prime factorization is useful in everyday life for tasks like finding the least common multiple of time intervals to schedule recurring events, or in computer science for data encryption algorithms. Similarly, working with exponents is fundamental in fields ranging from finance, where compound interest is calculated, to science, where we handle very large or small numbers using scientific notation. Mastering these operations helps in simplifying complex quantitative problems into manageable components.