Use logarithms to solve the given equation. (Round the answer to four decimal places.) HINT [See Example 1.]
3^(2x + 1) = 28

To solve the equation \( 3^{2x + 1} = 28 \) using logarithms, follow these steps:

### Step 1: Take the natural logarithm (ln) of both sides of the equation.
\[
\ln(3^{2x + 1}) = \ln(28)
\]

### Step 2: Apply the logarithmic power rule.
The logarithmic power rule states that \( \ln(a^b) = b \cdot \ln(a) \). Using this, we can move the exponent \( 2x + 1 \) in front of the logarithm:
\[
(2x + 1) \cdot \ln(3) = \ln(28)
\]

### Step 3: Solve for \( x \).
Now, distribute \( \ln(3) \) on the left-hand side:
\[
2x \cdot \ln(3) + 1 \cdot \ln(3) = \ln(28)
\]

Simplify further:
\[
2x \cdot \ln(3) + \ln(3) = \ln(28)
\]

Now isolate \( 2x \):
\[
2x \cdot \ln(3) = \ln(28) - \ln(3)
\]

Solve for \( x \):
\[
x = \frac{\ln(28) - \ln(3)}{2 \cdot \ln(3)}
\]

### Step 4: Calculate the logarithms.
Using a calculator:
- \( \ln(28) \approx 3.3322 \)
- \( \ln(3) \approx 1.0986 \)

Now substitute these values into the equation:
\[
x = \frac{3.3322 - 1.0986}{2 \cdot 1.0986}
\]
\[
x = \frac{2.2336}{2.1972}
\]
\[
x \approx 1.0166
\]

### Final Answer:
\( x \approx 1.0166 \) (rounded to four decimal places).

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Would you like more details or have any questions?

Here are five related questions to explore:
1. How would you solve \( 5^{3x - 4} = 125 \) using logarithms?
2. What is the logarithmic property that allows you to move exponents in front of the log?
3. Can you solve the equation \( 4^{x+2} = 50 \) using logarithms?
4. How would you rewrite the equation \( \log_b(x) = y \) in exponential form?
5. How can natural logarithms simplify equations involving powers?

**Tip**: Always use logarithms when trying to isolate variables in exponents!

Use logarithms to solve the given equation. (Round the answer to four decimal places.) 3^(2x + 1) = 28

Learn how to solve the exponential equation 3^(2x + 1) = 28 using logarithms. This solution covers logarithmic concepts, including the logarithmic power rule, and provides a detailed step-by-step explanation with final calculations. Suitable for students in Grades 9-12.

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