Give what appropriate subject this is in HighSchool Math (Algebra, Math Foundations, Algebra 2, Geometry, Integrated Math, Pre-Algebra, Calculus, PreCalculus, Trigonometry) Answer this according to guidelines (make it formal language. No first person POV): Solution Structure: Follow a step-by-step format: Introduction & Motivation (motivation example: Is there a property that helps us evaluate the limits of composite functions?), Theoretical Background, Application, Recap (as I have shown how), Result. Reformulate the question, introduce the most advanced concept, apply it, and state the result clearly. Style: Separate paragraphs with a blank line. Use default blue color for keywords and key concepts. Use italics to emphasize important words. Language: Use precise mathematical language and correct punctuation. Follow verb usage rules: “solve” for equations, “evaluate” for expressions. Equations: Use inline equations for simple expressions. Use display equations for complex formulas, centered separately. Use “$” before and after equations that are used inside or within a sentence Figures: Include captions, alt-text, and explanations directly with the figure. Use preferred colors and center all elements. Graphing Utility and Software: Use software only when required. Provide general instructions without mentioning the software’s name and ensure figures are original. Tables: Include a caption below each table, with explanations and conclusions in the same cell. Use the array environment and center the table and its caption (provide the actual result of the step by step solution and the answer and the KaTeX format for me to copy-paste) Question: Solve the equation and check for extraneous solutions.  $$log_{5}(x+4) + log_{5}(x+1) = 2$$

**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:**

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

---

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.

Solve the equation and check for extraneous solutions. $$log_{5}(x+4) + log_{5}(x+1) = 2$$

Learn how to solve the logarithmic equation log_5(x+4) + log_5(x+1) = 2 and check for extraneous solutions with a step-by-step guide. This problem covers key concepts in Algebra 2, including the logarithmic addition rule and quadratic factorization. Suitable for high school students in Algebra 2.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation