The problem is to solve the logarithmic equation:

$\log_2 3 + \log_2 x = \log_2 5 + \log_2 (x-2)$

Solution:

1. Combine logarithmic terms on both sides using the logarithmic property: $\log_b a + \log_b c = \log_b (a \cdot c)$

   Left-hand side: $\log_2 3 + \log_2 x = \log_2 (3x)$

   Right-hand side: $\log_2 5 + \log_2 (x - 2) = \log_2 [5(x - 2)]$

   The equation becomes: $\log_2 (3x) = \log_2 [5(x - 2)]$

2. Simplify logarithmic equality: $3x = 5(x - 2)$

3. Expand and solve for $x$: $3x = 5x - 10$ Subtract $3x$ from both sides: $0 = 2x - 10$ Add 10 to both sides: $10 = 2x$ Divide by 2: $x = 5$

4. Verify the solution: Substitute $x = 5$ into the original equation:

   • Left-hand side: $\log_2 3 + \log_2 5 = \log_2 (3 \cdot 5) = \log_2 15$
   • Right-hand side: $\log_2 5 + \log_2 (5 - 2) = \log_2 5 + \log_2 3 = \log_2 (5 \cdot 3) = \log_2 15$

   Both sides are equal, so $x = 5$ is the solution.

Final Answer:

$x = 5$

Would you like further clarification or more examples of logarithmic equations?

Related Questions:

1. What are the key properties of logarithms and how are they used in solving equations?
2. Can logarithmic equations have no solution? If so, under what conditions?
3. How do you graph logarithmic functions to verify solutions visually?
4. What happens if the domain restrictions of logarithms are not met?
5. How can you solve logarithmic equations with different bases?

Tip:

Always check the domain of the logarithmic equation to ensure that all arguments of the logarithms are positive. For example, here $x > 2$ is required because of $\log_2 (x-2)$.