What is the prime factorization using exponents of 63
Grammatically confusing question ..
I think you meant:
What is the prime factorization of 63, using exponents?
63 = 3x3x7
= 7x3^2
To find the prime factorization of 63 using exponents, we start by finding its prime factors:
1. Begin by dividing 63 by the smallest prime number, which is 2. However, 63 is not divisible by 2.
2. Next, try the next prime number, which is 3. Divide 63 by 3: 63 ÷ 3 = 21.
3. 21 is divisible by 3 without a remainder, so we continue dividing: 21 ÷ 3 = 7.
4. Finally, we reached a prime number, 7. As 7 is already a prime, we can't divide it further.
The prime factorization of 63 is therefore 3 x 3 x 7, which can be written as 3^2 x 7.
To find the prime factorization using exponents of 63, we need to break down the number into its prime factors.
Step 1: Start by dividing 63 by the smallest prime number, which is 2. Since 63 is odd, it is not divisible by 2.
Step 2: Move to the next prime number, which is 3. Dividing 63 by 3 gives us 21.
Step 3: The next prime number is also 3. Dividing 21 by 3 gives us 7.
Step 4: 7 is a prime number, so we cannot divide it further.
The prime factorization of 63 is therefore: 3 x 3 x 7.
To represent this in exponent form, we group like factors together:
63 = 3^2 x 7