Question

a $24 = 2^3 \times 3^1$
b $48 = 2^3 \times 6^1$
c $90 = 2^3 \times 5^1 \times 2^1$
d $68 = 34 \times 2$

Explanation:

Step1: Analyze Option A

Prime factorize 24: \(24 = 2\times12 = 2\times2\times6 = 2\times2\times2\times3 = 2^{3}\times3^{1}\). This is correct prime factorization.

Step2: Analyze Option B

Prime factors of 48: \(48=2\times24 = 2\times2\times12=2\times2\times2\times6 = 2\times2\times2\times2\times3=2^{4}\times3^{1}\). But option B has \(6^{1}\) and \(2^{3}\), and 6 is not a prime number, so this is incorrect.

Step3: Analyze Option C

Prime factorize 90: \(90 = 2\times45=2\times3\times15 = 2\times3\times3\times5=2^{1}\times3^{2}\times5^{1}\). Option C has incorrect exponents and repeated 2, so it's wrong.

Step4: Analyze Option D

Prime factorization requires prime numbers. 34 is not a prime number (\(34 = 2\times17\)), so \(68=2\times2\times17 = 2^{2}\times17^{1}\), and option D is not a prime factorization (just a product of two non - prime and prime), so it's incorrect.

Answer:

A. \(24 = 2^{3}\times3^{1}\)