Using the change-of-base formula, we write $\log_a b = \frac{\log b}{\log a}$: $\frac{\log 2}{\log 3} \cdot \frac{\log 3}{\log 4} \cdot \frac{\log 4}{\log 5} \dots \frac{\log (n-1)}{\log n} = \frac{1}{10}$.

Notice that the terms cancel out telescopically:

• The denominator $\log 3$ cancels with the next numerator $\log 3$.
• The denominator $\log 4$ cancels with the next numerator $\log 4$, and so on.

After all cancellations, only the first numerator and the last denominator remain: $\frac{\log 2}{\log n} = \frac{1}{10} \implies \log_n 2 = \frac{1}{10}$.

In exponential form, this is: $n^{1/10} = 2 \implies n = 2^{10} = 1024$.