Let $a_1,a_2,a_3,...$ be a sequence of positive real numbers such that:

(i) For all positive integers $m,n$, we have $a_{mn}=a_ma_n$

(ii) There exists a positive real number $B$ such that for all positive integers $m,n$ with $m<n$, we have $a_m < Ba_n$

Find all possible values of $\log_{2015}(a_{2015}) - \log_{2014}(a_{2014})$
1. **Given Conditions and Initial Observations:**
   - We are given a sequence of positive real numbers \(a_1, a_2, a_3, \ldots\) such that:
     \[
     a_{mn} = a_m a_n \quad \text{for all positive integers } m, n
     \]
     \[
     \exists B > 0 \text{ such that } a_m < B a_n \quad \text{for all positive integers } m, n \text{ with } m < n
     \]
   - We need to find the value of \(\log_{2015}(a_{2015}) - \log_{2014}(a_{2014})\).

2. **Sequence Property Analysis:**
   - From the first condition, we can deduce that \(a_{n^k} = a_n^k\) by repeated application of \(a_{mn} = a_m a_n\).
   - From the second condition, if \(m < n\), then:
     \[
     B > \frac{a_{m^k}}{a_{n^k}} = \left(\frac{a_m}{a_n}\right)^k
     \]
     This implies that \(\frac{a_m}{a_n} < B^{1/k}\) for any \(k\). As \(k \to \infty\), \(B^{1/k} \to 1\), so \(\frac{a_m}{a_n} \leq 1\). Hence, \(a_m \leq a_n\), proving that the sequence is non-decreasing.

3. **Determining \(a_1\):**
   - Since \(a_{1 \cdot 1} = a_1 \cdot a_1\), we have \(a_1 = a_1^2\). Given \(a_1\) is positive, it follows that \(a_1 = 1\).

4. **General Form of \(a_n\):**
   - We assert that \(a_n = n^c\) for some real number \(c\). To verify this, consider the prime factorization of \(n\):
     \[
     n = \prod_{i} p_i^{e_i}
     \]
     Using the property \(a_{mn} = a_m a_n\), we get:
     \[
     a_n = a_{\prod_{i} p_i^{e_i}} = \prod_{i} a_{p_i^{e_i}} = \prod_{i} (a_{p_i})^{e_i}
     \]
     Let \(a_{p_i} = p_i^c\). Then:
     \[
     a_n = \prod_{i} (p_i^c)^{e_i} = \left(\prod_{i} p_i^{e_i}\right)^c = n^c
     \]

5. **Verification of Conditions:**
   - The form \(a_n = n^c\) satisfies the first condition:
     \[
     a_{mn} = (mn)^c = m^c n^c = a_m a_n
     \]
   - For the second condition, we need \(a_m < B a_n\) for \(m < n\):
     \[
     \frac{a_m}{a_n} = \left(\frac{m}{n}\right)^c < B
     \]
     This holds for any \(c \geq 0\).

6. **Calculating the Desired Expression:**
   - Given \(a_n = n^c\), we have:
     \[
     \log_{2015}(a_{2015}) = \log_{2015}(2015^c) = c
     \]
     \[
     \log_{2014}(a_{2014}) = \log_{2014}(2014^c) = c
     \]
     Therefore:
     \[
     \log_{2015}(a_{2015}) - \log_{2014}(a_{2014}) = c - c = 0
     \]

The final answer is \(\boxed{0}\)