The equation provided is:

$2^{2x - 1} = 3^{2 - x}$

To solve for $x$, follow these steps:

Step 1: Take the natural logarithm (ln) of both sides

Apply the logarithm to both sides to make the exponents easier to handle:

$\ln(2^{2x - 1}) = \ln(3^{2 - x})$

Step 2: Use logarithmic properties

Using the property $\ln(a^b) = b \ln(a)$, we can rewrite the equation as:

$(2x - 1)\ln(2) = (2 - x)\ln(3)$

Step 3: Expand both sides

Distribute the logarithms:

$2x \ln(2) - \ln(2) = 2 \ln(3) - x \ln(3)$

Step 4: Group like terms

Collect all the terms involving $x$ on one side and constants on the other:

$2x \ln(2) + x \ln(3) = 2 \ln(3) + \ln(2)$

Factor out $x$:

$x(2 \ln(2) + \ln(3)) = 2 \ln(3) + \ln(2)$

Step 5: Solve for $x$

Finally, solve for $x$:

$x = \frac{2 \ln(3) + \ln(2)}{2 \ln(2) + \ln(3)}$

This is the exact value of $x$ in terms of natural logarithms. You can leave it in this form or compute an approximate value if necessary.

Would you like more details or explanations on any step?

Here are 5 follow-up questions to deepen understanding:

1. What logarithmic properties can simplify equations with exponents?
2. How can natural logarithms be applied to solve exponential equations?
3. What does it mean to find an "exact value" in logarithmic terms?
4. How does factoring help in solving equations?
5. Why are natural logarithms typically used in exponential equations?

Tip: Always apply logarithmic properties carefully when working with equations involving exponents.