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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.

Find the exact value of 2^(2x - 1) = 3^(2 - x)

This page explains how to solve the exponential equation 2^(2x - 1) = 3^(2 - x) using logarithmic properties. Follow a step-by-step approach to finding the exact value of x. Suitable for students in Grades 10-12.

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