We can see the reference solution jumps through several steps without proper explanation. Let us break this down so you can follow every move and understand the key concepts.

The key insight here is using the change of base formula: $\log_b x = \frac{\log x}{\log b}$ This lets us convert any logarithm to natural logs (or any common base). So our given terms become: $\log_{15} a = \frac{\log a}{\log 15}$ $\log_{32} a = \frac{\log a}{\log 32}$

Our equation: $\frac{\log_{15} a + \log_{32} a}{(\log_{15} a)(\log_{32} a)} = 4$ Becomes: $\frac{\frac{\log a}{\log 15} + \frac{\log a}{\log 32}}{\frac{\log a}{\log 15} \times \frac{\log a}{\log 32}} = 4$

$\frac{\log a}{\log 15} + \frac{\log a}{\log 32} = \log a \left(\frac{1}{\log 15} + \frac{1}{\log 32}\right)$ Using the common denominator rule: $= \log a \cdot \frac{\log 32 + \log 15}{\log 15 \cdot \log 32}$

$\frac{\log a}{\log 15} \times \frac{\log a}{\log 32} = \frac{(\log a)^2}{\log 15 \cdot \log 32}$

Our equation now looks like: $\frac{\log a \cdot \frac{\log 32 + \log 15}{\log 15 \cdot \log 32}}{\frac{(\log a)^2}{\log 15 \cdot \log 32}} = 4$ Here's the magic: When we divide fractions, we multiply by the reciprocal: $\frac{\log a \cdot (\log 32 + \log 15)}{(\log a)^2} = 4$ Notice how the $(\log 15 \cdot \log 32)$ terms cancel out!

$\frac{\log 32 + \log 15}{\log a} = 4$ Cross-multiplying: $\log 32 + \log 15 = 4 \log a$

Key Property: $\log x + \log y = \log(xy)$ and $n \log x = \log x^n$ So: $\log(32 \times 15) = \log a^4$ $\log 480 = \log a^4$ Therefore: $a^4 = 480$

We need $a = \sqrt[4]{480}$ Let's use perfect fourth powers to estimate: $4^4 = 256$ $5^4 = 625$ Since $256 < 480 < 625$, we have: $4 < \sqrt[4]{480} < 5$ Therefore: $4 < a < 5$

Key Takeaway: The change of base formula is your best friend when dealing with logarithms of different bases. Always look for opportunities to simplify complex logarithmic expressions by factoring out common terms!

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