Problem 28 - Intermediate Algebra

Question

Solve each equation. $$ 7^{x^{2}+3 x}=\frac{1}{49} $$

Step-by-Step Solution

Answer

The solutions are \( x = -1 \) and \( x = -2 \).

Step 1: Simplify the Right Side of the Equation

Recognize that the fraction \( \frac{1}{49} \) can be rewritten as a power of 7. Since \( 49 = 7^2 \), it follows that \( \frac{1}{49} = \frac{1}{7^2} = 7^{-2} \). Now, the equation is \( 7^{x^2 + 3x} = 7^{-2} \).

Step 2: Set Exponents Equal to Each Other

Since the bases are the same, we can set the exponents equal to each other: \( x^2 + 3x = -2 \). This results in a quadratic equation.

Step 3: Solve the Quadratic Equation

To solve the quadratic equation \( x^2 + 3x + 2 = 0 \), first factor it. The factors of 2 that add up to 3 are 1 and 2. So the equation factors to \((x + 1)(x + 2) = 0 \).

Step 4: Find the Solutions

Set each factor equal to zero and solve for \( x \). For \( x + 1 = 0 \), \( x = -1 \). For \( x + 2 = 0 \), \( x = -2 \).

Step 5: Check the Solutions

Substitute \( x = -1 \) and \( x = -2 \) back into \( 7^{x^2 + 3x} = 7^{-2} \) to verify. For \( x = -1 \), the left side becomes \( 7^{1 - 3} = 7^{-2} \). For \( x = -2 \), it becomes \( 7^{4 - 6} = 7^{-2} \). Both checks confirm the solutions.

Key Concepts

Exponent Rules

Exponent rules are essential in understanding how to manage expressions involving powers. In the given problem, recognizing how to manipulate exponents is crucial to simplify and solve the equation. Let's break this concept down:

When you have a fraction like \( \frac{1}{49} \), it can be converted into an exponential form if you understand the concept of negative exponents. The fraction \( \frac{1}{49} \) is equivalent to \( 49^{-1} \). Since \( 49 = 7^2 \), substituting gives us \( 7^{-2} \).

Here are some key points about exponent rules:

• Negative Exponents: These indicate the reciprocal of the base raised to the positive power. For instance, \( a^{-n} = \frac{1}{a^n} \).
• Same Bases: When equations have the same base, such as \( a^x = a^y \), the exponents must be equal: \( x = y \).

These rules allow us to rewrite problems in a simpler form, facilitating easier comparison and identification of solutions.

Factoring Polynomials

Factoring polynomials is a method used to simplify quadratic equations, enabling easier solution finding. When you encounter an equation like \( x^2 + 3x + 2 = 0 \), recognizing patterns is essential.

Here’s a simple approach to factoring such a polynomial:

1. Identify Necessary Factors: The goal is to find two numbers that multiply to the constant term (in this case, 2) and add up to the linear coefficient (in this case, 3).
2. Apply the Factors: In this scenario, 1 and 2 multiply to give 2 and add to 3, so they are our factors. Thus, we can express the polynomial as \((x + 1)(x + 2) = 0\).

This method transforms a complex quadratic equation into simpler linear equations \((x + 1) = 0\) and \((x + 2) = 0\), which are much easier to solve.

Verification of Solutions

Verification of solutions is the final step to ensure accuracy in problem-solving. It involves substituting the solution back into the original equation to check if it holds true. For this quadratic equation, let's consider the solutions \( x = -1 \) and \( x = -2 \):

• Substitute \( x = -1 \) into the left side of the equation: \( 7^{x^2 + 3x} \) becomes \( 7^{1 - 3} = 7^{-2} \), which matches the right side.
• For \( x = -2 \), the substitution gives \( 7^{4 - 6} = 7^{-2} \), again matching the right side.

Verification helps confirm each step was executed correctly and the derived solutions are valid. Always remember this crucial step to avoid unnecessary errors in the final answer.