Problem 28 - Elementary and Intermediate Algebra

Question

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

Step-by-Step Solution

Answer

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

Step 1: Identify Exponents

First, we need to rewrite the right side of the equation so that it has the same base as the left side. The left side is already expressed with base 7. Since 49 is equal to \(7^2\), we will rewrite \(\frac{1}{49}\) as \(7^{-2}\). Now the equation is \(7^{x^2 + 3x} = 7^{-2}\).

Step 2: Equate Exponents

Since the bases on both sides are the same (7), we can set the exponents equal to each other. Therefore, we have the equation \(x^2 + 3x = -2\).

Step 3: Rewrite as Quadratic Equation

Rearrange the equation to form a standard quadratic equation. Subtract -2 from both sides to get \(x^2 + 3x + 2 = 0\).

Step 4: Factor the Quadratic

Next, factor the quadratic equation. Look for two numbers that multiply to 2 and add to 3. These numbers are 1 and 2, so the equation factors into \((x+1)(x+2) = 0\).

Step 5: Solve for x

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

Step 6: Solution Conclusion

The solutions to the equation \(7^{x^2+3x} = \frac{1}{49}\) are \(x=-1\) and \(x=-2\).

Key Concepts

Understanding Quadratic Equations

Quadratic equations are fundamental in algebra and appear in various mathematical contexts. A quadratic equation is typically expressed in the standard form \(ax^2+bx+c=0\), where \(a\), \(b\), and \(c\) are constants, and \(x\) represents the variable. In our case, the equation \(x^2 + 3x + 2 = 0\) is a simple quadratic equation where:

• \(a = 1\)
• \(b = 3\)
• \(c = 2\)

Quadratic equations like this one often have two possible solutions because of the square term \(x^2\). The solutions are the values of \(x\) that make the equation true. Quadratic equations are typically solved using methods like factoring, completing the square, or using the quadratic formula. In this case, since the numbers involved are simple, factoring becomes a quick and efficient method.

The Power of Factoring

Factoring is breaking down complex expressions into simpler ones. For the quadratic equation, factoring involves finding two numbers whose product equals the constant term, \(c\), and whose sum equals the linear coefficient, \(b\). In our equation \(x^2 + 3x + 2\), we need to find two numbers that multiply to 2 and add to 3. These numbers are 1 and 2. Therefore, the quadratic expression can be factored as \((x+1)(x+2)\). Factoring is a crucial step because it simplifies the equation, allowing us to separate it into two distinct linear equations: \(x + 1 = 0\) and \(x + 2 = 0\). Solving these linear equations will give us the possible values of \(x\). Factoring not only makes solving quadratic equations easier but also helps in recognizing and understanding their structure.

Exploring Exponent Rules

Exponent rules are essential in algebra, especially when handling exponential equations like our original problem. Key rules allow us to simplify and manipulate expressions involving powers. Here are a few:

• When multiplying like bases, add the exponents: \(a^m \times a^n = a^{m+n}\).
• Inverse operations: \(a^{-n} = \frac{1}{a^n}\), which helps convert expressions to a consistent base.

For our equation \(7^{x^2 + 3x} = \frac{1}{49}\), utilizing these exponent rules allowed us to rewrite \(\frac{1}{49}\) as \(7^{-2}\), given that \(49 = 7^2\). This process of equating bases simplifies our work significantly, as it enables us to focus on the exponents themselves.

Understanding Base Equality

Base equality is a powerful concept in solving exponential equations. It relies on the principle that if two expressions with the same base are equal, then their exponents must be equal. This logic is used extensively to solve equations involving exponential terms. In our given exercise, we transformed \(\frac{1}{49}\) into \(7^{-2}\). With both sides of the equation now having the same base (7), we can equate the exponents for further simplification as: \[ x^2 + 3x = -2 \] By equating the exponents, we turned the problem into a quadratic equation, which is typically easier to solve. This step highlights the significance of expressing terms with a common base in algebra, simplifying complex exponential expressions into more manageable forms.