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The equation given is:

\[
\log_2 3 \times \log_3 4 \times \log_4 5 \times \dots \times \log_x (x+1) = 10
\]

To simplify this expression, we can use the **change of base property of logarithms**, which states:

\[
\log_a b = \frac{\log b}{\log a}
\]

By applying this rule to each term, we get:

\[
\log_2 3 = \frac{\log 3}{\log 2}, \quad \log_3 4 = \frac{\log 4}{\log 3}, \quad \log_4 5 = \frac{\log 5}{\log 4}, \dots, \quad \log_x (x+1) = \frac{\log (x+1)}{\log x}
\]

Now, the full product becomes:

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

Notice that most of the terms cancel out in a telescoping manner, leaving us with:

\[
\frac{\log (x+1)}{\log 2}
\]

Now, the equation simplifies to:

\[
\frac{\log (x+1)}{\log 2} = 10
\]

Multiplying both sides by \(\log 2\), we have:

\[
\log (x+1) = 10 \cdot \log 2
\]

Since \(\log 2\) is a constant (approximately \(0.3010\)), we can calculate:

\[
\log (x+1) = 10 \times 0.3010 = 3.010
\]

Now, exponentiating both sides (since we are dealing with a logarithmic equation), we get:

\[
x+1 = 10^{3.010}
\]

Calculating \(10^{3.010}\):

\[
x+1 \approx 1024
\]

Therefore:

\[
x \approx 1023
\]

Thus, the solution to the equation is \(x = 1023\).

---

Would you like more details on any step, or do you have any questions? 

Here are five related questions:
1. How does the change of base formula work for logarithms?
2. What is the meaning of a telescoping product?
3. How can logarithmic equations be solved by converting them to exponential form?
4. Why does the product of logarithms cancel out in this problem?
5. How can we use logarithmic properties to simplify more complex expressions?

**Tip**: Always look for patterns like telescoping in product or sum expressions to simplify logarithmic equations more easily.

log_2 3 × log_3 4 × log_4 5 × ... × log_x (x + 1) = 10

Learn how to solve a logarithmic product equation using the change of base formula and telescoping products. This problem covers advanced logarithmic concepts and exponential properties, making it ideal for students in Grades 11-12.

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