You can use the prime factorization of a number, written as the product of powers of prime numbers, to find the number of factors the number has. First, express the number as a product of powers of prime numbers; for example, 36=2^2 3^2 (that is 2 squared and 3 squared)Then increase each exponent by 1, and find the product of these increased exponents. This product is the number of factors of the number. In the example for 36, (2 + 1) (2 + 1) = 9, so 36 has 9 factors. Now determine how many factors 48 has.

What I have so far is the prime factorization is 48=2^4 x 3 and know that there are 10 factors but cannot figure out how to do the exponents thing listed above.

I wish they had explained this type of problem in a better way for you.

48 = 2x2x2x2x3

I can use the 2's in 5 ways, that is,
not take any
take 1
take 2
take 3
or take all 4

in the same way I can use the 3's in 2 ways,
take it or not take it

so that is 5x2 or 10 ways.

Be aware that this would include taking all the 2's and all the 3's, which results in 48 itself,
and the case of not taking any at all, which we could associate with the number 1, true in all cases.

Some purists would exclude 1 as a factor, so that way there would be only 9 factors.

5*2

To find the number of factors of 48 using the prime factorization, you are on the right track with expressing 48 as a product of powers of prime numbers.

You correctly identified that the prime factorization of 48 is 2^4 x 3.

Now, to apply the method described, you need to increase each exponent by 1 and multiply the increased exponents together.

For 48 = 2^4 x 3, increase the exponent of 2 by 1 to get 4 + 1 = 5. The exponent of 3 is already 1, so it remains the same.

Next, multiply the increased exponents together: 5 x 1 = 5.

Therefore, the number of factors of 48 is 5.

Let me know if you have any further questions or if there's anything else I can assist you with!