Logarithmic equation range for a

We can rewrite the given expression using base-change formula: $\frac{\log_{15} a + \log_{32} a}{(\log_{15} a)(\log_{32} a)} = \frac{1}{\log_a 15} + \frac{1}{\log_a 32} = \log_a 15 + \log_a 32 = \log_a (15 \times 32) = \log_a 480$

Given $\log_a 480 = 4 \implies a^4 = 480$.

Since $4^4 = 256$ and $5^4 = 625$, and $256 < 480 < 625$, we get: $4 < a < 5$.