To solve the exponential equation $2^{x-2} = 3^{2x}$, we will proceed step-by-step.

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Step 1: Rewrite the equation

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

Rewriting $2^{x-2}$ as $\frac{2^x}{2^2} = \frac{2^x}{4}$, the equation becomes: $\frac{2^x}{4} = 3^{2x}$

Multiply through by 4 to eliminate the denominator: $2^x = 4 \cdot 3^{2x}$

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Step 2: Take the natural logarithm of both sides

$\ln(2^x) = \ln(4 \cdot 3^{2x})$

Using logarithmic properties: $x \ln(2) = \ln(4) + \ln(3^{2x})$

$x \ln(2) = \ln(4) + 2x \ln(3)$

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Step 3: Rearrange to isolate $x$

Group terms involving $x$ on one side: $x \ln(2) - 2x \ln(3) = \ln(4)$

Factor out $x$: $x (\ln(2) - 2\ln(3)) = \ln(4)$

Solve for $x$: $x = \frac{\ln(4)}{\ln(2) - 2\ln(3)}$

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Step 4: Calculate the values

Using approximate values for logarithms:

• $\ln(2) \approx 0.693$
• $\ln(3) \approx 1.099$
• $\ln(4) = 2 \cdot \ln(2) \approx 1.386$

Substitute into the equation: $x = \frac{1.386}{0.693 - 2(1.099)}$

Simplify the denominator: $x = \frac{1.386}{0.693 - 2.198}$

$x = \frac{1.386}{-1.505}$

$x \approx -0.921$

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Final Answer:

The solution is: $x \approx -0.921$

Correct Choice: A. The solution set is $\{-0.921\}$.

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Do you want a deeper explanation or help with related problems? Here are 5 questions to expand your understanding:

1. How do logarithmic properties simplify solving exponential equations?
2. Why is taking the natural logarithm useful in this equation?
3. Can the equation $2^{x-2} = 3^{2x}$ have more than one solution?
4. How does the sign of the denominator affect the solution?
5. Could this problem be solved graphically, and what would that look like?

Tip: Always simplify the exponential base before taking logarithms; it can make calculations easier!