We need to solve the equation:

$\log_2 (x) = \log_x (2)$

Step 1: Express the equation in a common form

We know that the logarithm $\log_a(b)$ can be rewritten using the change of base formula:

$\log_a(b) = \frac{\log_c(b)}{\log_c(a)}$

Here, we will use the natural logarithm ($\ln$) for simplicity:

$\log_2(x) = \frac{\ln(x)}{\ln(2)}$

$\log_x(2) = \frac{\ln(2)}{\ln(x)}$

Substituting these into the original equation gives:

$\frac{\ln(x)}{\ln(2)} = \frac{\ln(2)}{\ln(x)}$

Step 2: Cross-multiply to solve for $x$

Cross-multiplying the equation:

$\ln(x) \times \ln(x) = \ln(2) \times \ln(2)$

This simplifies to:

$\ln^2(x) = \ln^2(2)$

Step 3: Take the square root of both sides

Taking the square root of both sides:

$\ln(x) = \pm \ln(2)$

Step 4: Solve for $x$

This gives us two cases:

1. $\ln(x) = \ln(2)$
   Taking the exponential of both sides: $x = 2$

2. $\ln(x) = -\ln(2)$
   Taking the exponential of both sides: $x = e^{-\ln(2)} = \frac{1}{2}$

Step 5: Check for extraneous solutions

We need to check if both solutions are valid. Let's substitute $x = 2$ and $x = \frac{1}{2}$ back into the original equation:

• For $x = 2$:
  $\log_2(2) = 1 \quad \text{and} \quad \log_2(2) = 1$ So, $x = 2$ is a valid solution.

• For $x = \frac{1}{2}$:
  $\log_2\left(\frac{1}{2}\right) = -1 \quad \text{and} \quad \log_{\frac{1}{2}}(2) = -1$ So, $x = \frac{1}{2}$ is also a valid solution.

Final Answer:

The solutions to the equation are:

$x = 2 \quad \text{or} \quad x = \frac{1}{2}$

Would you like more details or have any questions?

---

Here are 5 related questions to explore:

1. What is the solution to $\log_3(x) = \log_x(3)$?
2. Solve $\log_5(x) = \log_x(5)$.
3. What happens if the base of the logarithms is a fraction, like $\log_{\frac{1}{3}}(x) = \log_x\left(\frac{1}{3}\right)$?
4. Can you solve $\log_a(x) = \log_x(a)$ for a general positive $a$?
5. How do you solve $\log_x(x^2) = \log_2(4)$?

Tip: When dealing with logarithmic equations, always check for extraneous solutions that might arise from the logarithm's domain constraints.