To solve the equation \( \log(7x + 1) = 2\log(x + 3) - \log 2 \), follow these steps:

1. **Simplify the equation using logarithm properties:**

   First, use the properties of logarithms to combine the terms on the right-hand side. The properties we will use are:
   \[
   a \log b = \log b^a \quad \text{and} \quad \log a - \log b = \log \left(\frac{a}{b}\right)
   \]

   Rewrite the right-hand side:
   \[
   2\log(x + 3) - \log 2 = \log((x + 3)^2) - \log 2 = \log\left(\frac{(x + 3)^2}{2}\right)
   \]

   Thus, the equation becomes:
   \[
   \log(7x + 1) = \log\left(\frac{(x + 3)^2}{2}\right)
   \]

2. **Remove the logarithms:**

   Since the logarithms are equal, their arguments must be equal:
   \[
   7x + 1 = \frac{(x + 3)^2}{2}
   \]

3. **Clear the fraction:**

   Multiply both sides by 2 to get rid of the denominator:
   \[
   2(7x + 1) = (x + 3)^2
   \]
   \[
   14x + 2 = (x + 3)^2
   \]

4. **Expand and simplify:**

   Expand the right-hand side:
   \[
   14x + 2 = x^2 + 6x + 9
   \]

   Move all terms to one side to form a quadratic equation:
   \[
   x^2 + 6x + 9 - 14x - 2 = 0
   \]
   \[
   x^2 - 8x + 7 = 0
   \]

5. **Solve the quadratic equation:**

   Factor the quadratic equation:
   \[
   x^2 - 8x + 7 = (x - 1)(x - 7) = 0
   \]

   Set each factor equal to zero and solve for \( x \):
   \[
   x - 1 = 0 \quad \text{or} \quad x - 7 = 0
   \]
   \[
   x = 1 \quad \text{or} \quad x = 7
   \]

6. **Check for extraneous solutions:**

   Substitute \( x = 1 \) back into the original equation:
   \[
   \log(7(1) + 1) = 2\log(1 + 3) - \log 2
   \]
   \[
   \log 8 = 2\log 4 - \log 2
   \]
   \[
   \log 8 = \log 16 - \log 2
   \]
   \[
   \log 8 = \log \left(\frac{16}{2}\right)
   \]
   \[
   \log 8 = \log 8 \quad \text{(True)}
   \]

   Substitute \( x = 7 \) back into the original equation:
   \[
   \log(7(7) + 1) = 2\log(7 + 3) - \log 2
   \]
   \[
   \log 50 = 2\log 10 - \log 2
   \]
   \[
   \log 50 = \log 100 - \log 2
   \]
   \[
   \log 50 = \log \left(\frac{100}{2}\right)
   \]
   \[
   \log 50 = \log 50 \quad \text{(True)}
   \]

Both solutions \( x = 1 \) and \( x = 7 \) are valid.

So, the solutions to the equation are:
\[
x = 1 \quad \text{and} \quad x = 7
\]

Would you like more details or have any questions?

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Here are 8 related questions to expand your understanding:
1. How do you solve equations involving natural logarithms (\(\ln\))?
2. What are the properties of logarithms and how are they applied in solving equations?
3. How do you solve exponential equations using logarithms?
4. What is the change of base formula for logarithms?
5. How can you solve a system of equations involving logarithms?
6. How do you graph logarithmic functions and their transformations?
7. What are some real-life applications of logarithmic functions?
8. How do you solve logarithmic inequalities?

**Tip:** When solving logarithmic equations, always check for extraneous solutions by substituting back into the original equation.

Learn how to solve the logarithmic equation log(7x + 1) = 2log(x + 3) - log 2 with detailed step-by-step solutions and verification of solutions. This problem covers logarithmic properties, quadratic equations, and their applications. Suitable for high school students in Grades 10-12.

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