Introduction & Motivation:

In logarithmic equations, certain properties allow us to combine and simplify expressions. One such property is the logarithmic addition rule:

$\log_b(A) + \log_b(B) = \log_b(A \cdot B)$

This property will help us combine the logarithms on the left-hand side of the equation and solve for the unknown variable.

Theoretical Background:

The given equation is:

$\log_5(x+4) + \log_5(x+1) = 2$

We can apply the logarithmic addition rule to the left-hand side. According to this rule, when adding two logarithms with the same base, we can combine them into one logarithm:

$\log_5((x+4)(x+1)) = 2$

Once the logarithms are combined, we can eliminate the logarithmic function by exponentiating both sides with the base of 5. This transforms the equation into a quadratic form, which can be solved algebraically.

Application:

1. Combine the logarithmic expressions:

   Using the logarithmic addition rule:

   $\log_5((x+4)(x+1)) = 2$

2. Exponentiate both sides:

   Since the base of the logarithm is 5, we raise 5 to the power of both sides to eliminate the logarithm:

   $(x+4)(x+1) = 5^2$

   Simplifying:

   $(x+4)(x+1) = 25$

3. Expand the product:

   Expanding the left-hand side:

   $x^2 + 5x + 4 = 25$

4. Form the quadratic equation:

   Subtract 25 from both sides to set the equation equal to zero:

   $x^2 + 5x - 21 = 0$

5. Solve the quadratic equation:

   To solve this quadratic equation, we use factoring:

   $(x + 7)(x - 3) = 0$

   Therefore, the solutions are:

   $x = -7 \quad \text{or} \quad x = 3$

Recap:

After solving the quadratic equation, we found two potential solutions: $x = -7$ and $x = 3$. However, logarithmic functions have domain restrictions; the arguments inside the logarithms must be positive. Therefore, we need to check for extraneous solutions.

Check for Extraneous Solutions:

1. Substitute $x = -7$ into the original logarithmic expressions:

   $\log_5(-7+4) + \log_5(-7+1) = \log_5(-3) + \log_5(-6)$

   Both expressions involve the logarithm of a negative number, which is undefined. Thus, $x = -7$ is an extraneous solution.

2. Substitute $x = 3$ into the original logarithmic expressions:

   $\log_5(3+4) + \log_5(3+1) = \log_5(7) + \log_5(4)$

   Both terms are defined and positive. Thus, $x = 3$ is a valid solution.

Result:

The solution to the equation is:

$x = 3$

The extraneous solution $x = -7$ is discarded due to domain restrictions.

KaTeX Format:

$\log_5(x+4) + \log_5(x+1) = 2$ $\log_5((x+4)(x+1)) = 2$ $(x+4)(x+1) = 25$ $x^2 + 5x - 21 = 0$ $(x+7)(x-3) = 0$ $x = 3$

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Would you like further details or clarification? Here are five related questions:

1. How can logarithmic equations be solved when the bases are different?
2. What is an extraneous solution, and why does it occur in logarithmic equations?
3. How can we apply the quadratic formula to solve non-factorable quadratic equations?
4. What are the domain restrictions for logarithmic functions, and how do they influence solutions?
5. How does the base of a logarithm affect the solution of logarithmic equations?

Tip: Always check the domain of the logarithmic functions before finalizing the solution, as extraneous solutions can often arise from restrictions on logarithmic inputs.