https://file.aiquickdraw.com/mathbot/file/17307951810986mfq9vgq.jpg

The problem in the image reads:

**Solve \( \log_2 3 \times \log_3 4 \times \log_4 5 \times \dots \times \log_{2n-1} (2n) = 5 \), where \( n \) is an integer.**

### Solution

Notice that each term in the product is of the form \( \log_k (k+1) \). Using the change of base formula, we can rewrite each term \( \log_k (k+1) \) as follows:

\[
\log_k (k+1) = \frac{\log (k+1)}{\log k}
\]

So we have:
\[
\log_2 3 \times \log_3 4 \times \log_4 5 \times \dots \times \log_{2n-1} (2n)
\]

This product of logarithms can be rewritten as:

\[
\frac{\log 3}{\log 2} \times \frac{\log 4}{\log 3} \times \frac{\log 5}{\log 4} \times \dots \times \frac{\log (2n)}{\log (2n - 1)}
\]

Observe that in this sequence, all the terms cancel out except for the very first and the very last terms. So we get:

\[
\frac{\log (2n)}{\log 2} = \log_2 (2n)
\]

Thus, the original product simplifies to:

\[
\log_2 (2n) = 5
\]

Now, we solve for \( n \) by rewriting the equation in exponential form:

\[
2n = 2^5
\]

\[
2n = 32
\]

\[
n = 16
\]

### Conclusion
The solution is:
\[
n = 16
\]

Would you like further clarification, or do you have any additional questions?

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### Related Questions

1. How does the change of base formula work in logarithmic expressions?
2. What are some other ways to simplify products of logarithmic terms?
3. How does this problem change if we adjust the base of the logarithms?
4. What if the problem asked for the value of \( n \) such that \( \log_2 (2n) = k \) for an arbitrary integer \( k \)?
5. Can the telescoping property be applied in different forms of sequences?

**Tip:** Recognizing patterns in logarithmic sequences can often simplify complex products significantly. Look for ways terms might cancel each other out!

Solve log_2 3 × log_3 4 × log_4 5 × ... × log_(2n-1) (2n) = 5, where n is an integer.

Explore the solution to the logarithmic product equation log_2 3 × log_3 4 × ... × log_(2n-1) (2n) = 5, using the telescoping property and algebraic simplification. Understand key concepts like the change of base formula and telescoping products, suitable for advanced high school math students.

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