We are tasked with solving the equation:

$\log_7(4x) - \log_7(x + 1) = 1 + 2\log_7(4).$

Step-by-step Solution

1. Simplify the right-hand side:

Using the logarithmic property $a\log_b(c) = \log_b(c^a)$, rewrite $2\log_7(4)$ as: $2\log_7(4) = \log_7(4^2) = \log_7(16).$ Thus, the equation becomes: $\log_7(4x) - \log_7(x + 1) = 1 + \log_7(16).$

2. Combine terms on the right-hand side:

Using the property $\log_b(A) + \log_b(B) = \log_b(A \cdot B)$, rewrite $1 + \log_7(16)$ as: $1 = \log_7(7), \quad \therefore 1 + \log_7(16) = \log_7(7 \cdot 16) = \log_7(112).$ Thus, the equation becomes: $\log_7(4x) - \log_7(x + 1) = \log_7(112).$

3. Simplify the left-hand side:

Using the property $\log_b(A) - \log_b(B) = \log_b\left(\frac{A}{B}\right)$, rewrite the left-hand side as: $\log_7\left(\frac{4x}{x + 1}\right).$ Thus, the equation is now: $\log_7\left(\frac{4x}{x + 1}\right) = \log_7(112).$

4. Remove the logarithms:

Since the equation $\log_b(A) = \log_b(B)$ implies $A = B$, we get: $\frac{4x}{x + 1} = 112.$

5. Solve for $x$:

Multiply through by $x + 1$ (assuming $x \neq -1$): $4x = 112(x + 1).$ Expand the right-hand side: $4x = 112x + 112.$ Rearrange terms: $4x - 112x = 112 \quad \implies \quad -108x = 112.$ Solve for $x$: $x = -\frac{112}{108} = -\frac{28}{27}.$

6. Verify the solution:

Since $\log(x)$ is undefined for non-positive $x$, the solution $x = -\frac{28}{27}$ is not valid.

Conclusion:

There is no solution to the given equation.

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Do you have questions about any part of the solution, or would you like further clarification?

Relative Questions:

1. What are the properties of logarithms used in this problem?
2. Can logarithmic equations have no solution, and why?
3. How does the domain of logarithmic functions affect solutions?
4. What would happen if $x$ had been positive in this case?
5. How can you graphically check for solutions to this equation?

Tip:

Always verify the domain of the logarithmic terms to ensure the solutions are valid within the equation's constraints!