Let n be a natural number greater than 1 and suppose n has k distinct prime factors. Please prove that log(n) ≥ k log(2).

**Solution. **Congrats to Sean Pegado, the first to submit the following correct solution:

If n = p_1^e_1 * p_2^e_2 * …. * p^e_k, we know that the smallest possible value for p_i will be 2. Since we have distinct prime factors, we can rewrite

n = p_1^e_1 * p_2^e_2 * …. * p^e_k ≥ 2^e_1 * … * 2^e_k = 2^{e_1 + … + e_k}.

Since we have k distinct prime factors, we know that each e_i must be positive, and thus greater than or equal to 1.

n ≥ 2^{e_1 + … + e_k} ≥ 2^{1+…+1} = 2^k.

So n ≥ 2^k, and log(n) ≥ k*log(2).