The given equation is:

$\log_2 3 \times \log_3 4 \times \log_4 5 \times \cdots \times \log_x (x+1) = 10$

Step-by-Step Solution:

This type of equation involves a product of logarithms with bases and arguments progressing incrementally. We can utilize the change of base formula for logarithms to simplify the expression.

Step 1: Recognize the general pattern.

Each term in the product can be simplified using the change of base formula: $\log_b a = \frac{\log a}{\log b}$ This allows us to rewrite each term in the product as: $\log_2 3 = \frac{\log 3}{\log 2}, \quad \log_3 4 = \frac{\log 4}{\log 3}, \quad \log_4 5 = \frac{\log 5}{\log 4}, \quad \dots, \quad \log_x (x+1) = \frac{\log(x+1)}{\log x}$

Step 2: General simplification.

The product of these logarithms telescopes. This is because for each adjacent logarithm, the numerator of one term is canceled by the denominator of the previous term: $\frac{\log 3}{\log 2} \times \frac{\log 4}{\log 3} \times \frac{\log 5}{\log 4} \times \cdots \times \frac{\log(x+1)}{\log x} = \frac{\log(x+1)}{\log 2}$

Step 3: Set up the equation.

Now, the equation reduces to: $\frac{\log(x+1)}{\log 2} = 10$

Step 4: Solve for $x$.

Multiply both sides by $\log 2$ to isolate the logarithm: $\log(x+1) = 10 \times \log 2$ $\log(x+1) = \log(2^{10})$ $\log(x+1) = \log 1024$

Since the logarithms are equal, their arguments must be equal: $x + 1 = 1024$ $x = 1023$

Thus, the solution to the equation is $x = 1023$.

Let me know if you want more details or have any questions.

Here are 5 related questions for further exploration:

1. How does the change of base formula work in logarithms?
2. What is a telescoping product and how does it simplify this type of equation?
3. Can this problem be generalized to different logarithmic products?
4. What would happen if the product had a finite or infinite number of terms?
5. What is the role of exponential growth in logarithmic equations?

Tip: The change of base formula is a crucial tool when dealing with logarithms of different bases. Always check if adjacent terms can telescope for simplification.