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To solve the equation \( 2(3^{2x-1}) = 4^{x+1} \), we can follow these steps:

1. Express both sides with the same base. Note that \( 4 = 2^2 \):

\[ 4^{x+1} = (2^2)^{x+1} = 2^{2(x+1)} = 2^{2x+2} \]

So the equation becomes:

\[ 2(3^{2x-1}) = 2^{2x+2} \]

2. Divide both sides by 2:

\[ 3^{2x-1} = 2^{2x+1} \]

3. Take the natural logarithm (ln) of both sides to remove the exponents:

\[ \ln(3^{2x-1}) = \ln(2^{2x+1}) \]

Using the logarithm power rule \( \ln(a^b) = b \ln(a) \):

\[ (2x-1) \ln(3) = (2x+1) \ln(2) \]

4. Expand and rearrange the equation to isolate \( x \):

\[ 2x \ln(3) - \ln(3) = 2x \ln(2) + \ln(2) \]

\[ 2x \ln(3) - 2x \ln(2) = \ln(3) + \ln(2) \]

\[ 2x (\ln(3) - \ln(2)) = \ln(3) + \ln(2) \]

\[ x = \frac{\ln(3) + \ln(2)}{2(\ln(3) - \ln(2))} \]

5. Calculate the value using a calculator:

\[ x = \frac{\ln(3) + \ln(2)}{2(\ln(3) - \ln(2))} \]

Using the natural logarithm values:
- \( \ln(3) \approx 1.0986 \)
- \( \ln(2) \approx 0.6931 \)

\[ x = \frac{1.0986 + 0.6931}{2(1.0986 - 0.6931)} \]

\[ x = \frac{1.7917}{2 \times 0.4055} \]

\[ x = \frac{1.7917}{0.8110} \]

\[ x \approx 2.21 \]

Therefore, the solution to the equation \( 2(3^{2x-1}) = 4^{x+1} \) is \( x \approx 2.21 \).

Would you like any further details or have any questions?

Here are five related questions:
1. How can we use logarithms to solve exponential equations in general?
2. What is the logarithm power rule, and how is it applied?
3. Can you explain the steps to isolate \( x \) in an equation involving logarithms?
4. How do we convert a base in an exponential equation to simplify the problem?
5. What are some other methods to solve exponential equations besides using logarithms?

**Tip:** Always check your solutions by substituting them back into the original equation to ensure they satisfy it.

Solve the equation 2(3^{2x-1}) = 4^{x+1}, giving your answer correct to 2 decimal places.

This page provides a step-by-step solution to the exponential equation 2(3^{2x-1}) = 4^{x+1}. The solution involves using logarithms and algebra to simplify and solve the equation, providing the answer to two decimal places. Ideal for students in Grades 9-12.

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