Problem 34 - Algebra and Trigonometry

Question

Solve the equation. \(x^{2} 10^{x}-x 10^{x}=2\left(10^{x}\right)\)

Step-by-Step Solution

Answer

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

Step 1: Factor Out Common Terms

We start by factoring out the common term \(10^x\) from each term on the left side of the equation: \[ x^2 \cdot 10^x - x \cdot 10^x = 2 \cdot 10^x \]This simplifies to:\[ 10^x (x^2 - x) = 2 \cdot 10^x \]

Step 2: Simplify the Equation

Since \(10^x\) is on both sides of the equation and cannot be zero (because \(10^x\) is positive for all real numbers \(x\)), we can divide both sides by \(10^x\) to simplify:\[ x^2 - x = 2 \]

Step 3: Rearrange into Quadratic Form

Rearrange the simplified equation into a standard quadratic equation:\[ x^2 - x - 2 = 0 \]

Step 4: Solve the Quadratic Equation

We solve the quadratic equation \(x^2 - x - 2 = 0\) by factoring it. We look for two numbers that multiply to \(-2\) and add to \(-1\): The numbers \(-2\) and \(1\) satisfy both conditions. Therefore, we factor the quadratic as:\[ (x - 2)(x + 1) = 0 \]

Step 5: Find the Solution

Setting each factor equal to zero, we solve for \(x\):1. \(x - 2 = 0\) - Solving this gives \(x = 2\).2. \(x + 1 = 0\) - Solving this gives \(x = -1\). The solutions are \(x = 2\) and \(x = -1\).

Key Concepts

Factoring Quadratics

Factoring quadratic equations can seem daunting at first, but with practice, it becomes second nature. Typically, we want to express a quadratic equation of the form \(ax^2 + bx + c = 0\) in a way that allows us to find its roots. This involves turning it into a product of two binomials, like

• \((x - p)(x - q) = 0\)

Here, \(p\) and \(q\) are the roots of the equation.
The process of factoring involves finding two numbers that multiply to \(c\) (the constant term) and add up to \(b\) (the coefficient of the linear term \(x\)). Let's take our equation \(x^2 - x - 2 = 0\). We need numbers that multiply to \(-2\) and add to \(-1\). Recognizing these numbers as \(-2\) and \(1\), we rewrite the equation as

• \((x - 2)(x + 1)\)

Setting each factor to zero gives us the solutions: \(x - 2 = 0\) gives \(x = 2\) and \(x + 1 = 0\) gives \(x = -1\). Remember, learning to factor efficiently is vital in solving many quadratic equations.

Exponential Functions

Exponential functions feature prominently in mathematics due to their unique growth patterns. They are of the form \(a^x\), where \(a\) is a constant and \(x\) is an exponent. The function \(10^x\) in our exercise is an example of an exponential function, where 10 is the base number. Such functions grow rapidly as \(x\) increases.
The function \(10^x\) is always positive, which is a critical property utilized during the equation solving process.

• For instance, since \(10^x > 0\) for all \(x\), we knew it was safe to divide by \(10^x\) on both sides of the equation \(10^x(x^2 - x) = 2 \cdot 10^x\) without worrying about dividing by zero.
• Moreover, this property of exponential functions enables simplifications, particularly when the function appears as a common factor in equations.

Understanding how these functions behave helps in solving complex equations more easily.

Equation Solving Steps

Solving equations methodically ensures clarity and precision. Here, we tackled the equation \(x^2 \cdot 10^x - x \cdot 10^x = 2 \cdot 10^x\) by following structured steps.
**Step 1: Factor Common Terms**

• We noticed \(10^x\) appeared in all terms, allowing us to factor it out, drastically simplifying the process.

**Step 2: Eliminate the Common Factor**

• Since \(10^x\) is non-zero for any real \(x\), we safely divided the entire equation by \(10^x\), focusing solely on the quadratic part.

**Step 3: Quadratic Transformation**

• After simplifying, we were left with a standard quadratic form: \(x^2 - x - 2 = 0\), easier to solve.

**Step 4: Solve the Quadratic Equation**

• Factoring this quadratic provided us with solutions \(x = 2\) and \(x = -1\).

Following these steps not only streamlines the solving process but also enhances understanding of how various mathematical components intertwine.