For public/private key cryptography, you need to generate a private key, and from this private key you can create or derive your public key. In the case of RSA, you generate a public key in the following manner:
1.) Choose two unique, large prime numbers. (p and q)
2.) Multiply these unique primes together (result = n)
3.) Compute: (p-1)*(q-1) (result = φ)
4.) Pick an integer between 1 and φ which doesn't share a prime factor with φ (result = e)
5.) Figure out what number (d) when multiplied by (e/φ) gives 1. In other words, the remainder of (d*e)/φ is 1.

So, let's do an example! :)
1.) My two primes (obviously waaaay to small to ever be practical for actual cryptography): p=7 and q=13
2.) Multiply them(n=91)
3.) Multiply (p-1) with (q-1): (6*12)φ=72
4.) Pick an integer sharing no prime factor with φ:e=11
5.) Number which when multiplied by (e/φ) gives us a remainder of one:d=51
-->Check: d*(e/φ)=649/72=9 + 1/72 or 9 remainder 1. YAY!

Now, your public key is the two numbers 'e' and 'n', so our public key is (11, 91). 
Your private key is the two numbers 'd' and 'n', so our private key is (51, 91). 

Now, of course these numbers are ridiculously small and not practical for encryption, but let's demonstrate breaking this simple RSA key given the availability of a 'function' B(x) which, when fed a number, would return the prime factorization. Example: B(391) returns two numbers: 17 and 23. THIS FUNCTION B(x) is what quantum computers would provide.

Given only the public key, we have two variables initially:
e = 11
n = 91
As you can see above, if we could calculate both p and q (the initial starting prime numbers), we could re-calculate the private key d. Given the starting primes, finding 'd' is deterministic. So, our only goal here is to calculate p and q. We already know pq, and we know p and q are both primes. To this effect, calling B(n) or B(91) yields {7, 13}. However, if we were unable to easily find the prime factorization of 91, we would be stuck brute-forcing:
2*x=91 (x not integer)
3*x=91 (x not integer)
5*x=91 (x not integer)
7*x=91 (x is an integer)
For very large values of  n, brute-forcing becomes computationally-infeasible. Another means of attacking RSA keys is the number field sieve, which is significantly more effective, and was used to crack a 768-bit RSA key. For a key of size 3072-bit, it would take all the computing power on Earth today longer (by orders of magnitude) than the age of the known universe to factor this number 'e' back into parts 'p' and 'q'. 

Or, with a quantum computer of large enough (around 6144 'qubits' or quantum bits, a unit of quantum information which can exist in a superposition where it is both 1 AND 0), you'd just use Shor's algorithm (which has been demonstrated in factoring '15' into '3' and '5' (not impressive in itself of course, but the technology exists and Shor's algorithm is provably feasible) to call B(n). 

The discrete logarithm problem is another mathematical entity believed to be easy one way, hard the other. 
For example:
q^b mod n ≡ x
Where n is a primitive root of q (basically, for all possible values of r, the chances of getting any result (1, 2, 3, 4, 5, 6) from (q^b MOD n) is equal)
Solve for x. 
Knowing n, q, and b, this can be calculated very easily. Let's say n = 13. 6 is a primitive root. 
How we know 6 is a primitive root of 13:
6^1 MOD 13 = 6
6^2 MOD 13 = 10
6^3 MOD 13 = 8
6^4 MOD 13 = 9
6^5 MOD 13 = 2
6^6 MOD 13 = 12
6^7 MOD 13 = 7
6^8 MOD 13 = 3
6^9 MOD 13 = 5
6^10 MOD 13 = 4
6^11 MOD 13 = 11
6^12 MOD 13 = 1
6^13 MOD 13 = 6
6^14 MOD 13 = 10
6^15 MOD 13 = 8
etc... You can see a pattern. 6^1 MOD 13 is the same as 6^13 MOD 13 is the same as 6^25 MOD 13 is the same as 6^37 MOD 13. Using this looping pattern, we can predict 6^1994833 MOD 6 as being 6. Why? Because (1994833-1)/12 is an integer. We don't have to calculate 6^1994833, we just use a pattern. Cool, huh? 6^1994834 MOD 13 is 10. 6^1994835 MOD 13 is 8. Etc. Not really important to what we're doing right now, it's just awesome. :D

Aaaanyhow, back to what we were doing. 
 q^b mod n ≡ x
Let's set q as 6, n as 13, and b as 119. We could either use our pattern above or resort to asking a calculator, but x is easy to calculate. x = 11.
That's easy. Now, try calculating:
6^b mod 13 = 119. 
For such a small modulus 13, this problem is trivial trial-and error. However, this problem becomes far more difficult with extremely large moduli, similar to a prime factorization.