Problem 26 - Intermediate Algebra

Question

Solve each equation. $$ 3^{x^{2}-3 x}=81 $$

Step-by-Step Solution

Answer

The solutions are \(x = 4\) and \(x = -1\).

Step 1: Express 81 as a Power of 3

First, recognize that 81 is a power of 3. We can write 81 as a power of 3: 81 = \(3^4\). This helps in equating the powers of 3 on both sides of the equation, which simplifies the process of solving for \(x\).

Step 2: Set the Exponents Equal

Since the bases are the same (both sides of the equation have base 3), we can set the exponents equal to each other. This gives us the equation \(x^2 - 3x = 4\).

Step 3: Rearrange into a Quadratic Equation

We rearrange the equation to form a standard quadratic equation: \(x^2 - 3x - 4 = 0\). Now, our task is to solve this quadratic equation.

Step 4: Solve the Quadratic Equation

We factor the quadratic equation \(x^2 - 3x - 4 = 0\). Look for two numbers that multiply to -4 and add to -3. The numbers are -4 and 1. Thus, the equation factors to \((x - 4)(x + 1) = 0\).

Step 5: Find the Roots of the Factors

Set each factor equal to zero and solve the equations.\[(x - 4) = 0 \Rightarrow x = 4\]\[(x + 1) = 0 \Rightarrow x = -1\] Therefore, the solutions to the equation are \(x = 4\) and \(x = -1\).

Key Concepts

Quadratic Equations

An essential part of solving equations like the one above involves understanding quadratic equations. A quadratic equation is a polynomial equation of degree 2. It generally takes the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants. In our problem, the equation \( x^2 - 3x - 4 = 0 \) fits this form. Quadratic equations have a characteristic 'U-shape' when plotted on a graph, known as a parabola. The solutions or roots of a quadratic equation are the points where this parabola crosses the x-axis.

There are several methods to solve quadratic equations:

• Factoring,
• Using the quadratic formula,
• Completing the square.

In our specific problem, factoring was the simplest method to use. Once the quadratic equation is rewritten in the factored form, finding the solutions involves setting each factor equal to zero and solving for \( x \). This is because if a product of two numbers is zero, then at least one of the numbers must be zero.

Factoring

Factoring is a powerful tool in mathematics used to break down expressions into simpler, more manageable parts. It is especially useful for solving equations, such as quadratics. In the context of our example, we factored the quadratic \( x^2 - 3x - 4 = 0 \) into the product \((x - 4)(x + 1)\). This means we searched for two numbers that multiply to \(-4\) (the constant term) and add to \(-3\) (the linear coefficient).

• The two numbers are \(-4\) and \(1\).
• These numbers were chosen because \(-4 \times 1 = -4\) and \(-4 + 1 = -3\).

By factoring, the complicated process of solving a quadratic reduces to simple arithmetic. You then find the values of \( x \) that make each factor zero: \((x - 4) = 0\) or \((x + 1) = 0\), leading to \(x = 4\) and \(x = -1\).

Factoring not only gives us the roots of the quadratic but also provides insight into the symmetry of the parabola and its intersection points with the x-axis.

Exponential and Logarithmic Functions

Exponential and logarithmic functions are closely related and essential in solving equations where the variable is an exponent, such as in our original exercise \(3^{x^2 - 3x} = 81\). Recognizing the number 81 as a power of 3, specifically \(3^4\), was a key step. This converted the problem into one involving equal bases: \(3^{x^2 - 3x} = 3^4\), allowing us to equate the exponents directly.

Here’s why understanding these functions is crucial:

• **Exponential Functions** involve equations where the variable is in the exponent, often written in the form \(b^x\).
• **Logarithmic Functions** are the inverses of exponential functions and are used to solve for the exponent: if \(b^y = x\), then \(y = \log_b(x)\).

The process simplifies dramatically if both sides of an equation can be rewritten with the same base, enabling a straightforward comparison of exponents. This is what we did by expressing both sides in terms of base 3.

Understanding the interplay between exponential and logarithmic functions allows for efficient solving of complex equations, making them indispensable tools in the mathematician’s arsenal.