Problem 50 - Precalculus with Calculus Previews

Question

Either use factoring or the quadratic formula to solve the given equation. $$ 10^{2 x}-103\left(10^{x}\right)+300=0 $$

Step-by-Step Solution

Answer

The solutions for x are 2 and \( \log_{10}(3) \).

Step 1: Recognize the Structure of the Equation

The given equation is a quadratic form in terms of \( y = 10^x \). Thus, the equation becomes \( y^2 - 103y + 300 = 0 \).

Step 2: Apply the Quadratic Formula

The quadratic formula is given by \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1 \), \( b = -103 \), and \( c = 300 \). Plug these values into the formula to find \( y \).

Step 3: Calculate the Discriminant

Calculate the discriminant \( b^2 - 4ac \) as follows: \( (-103)^2 - 4 \times 1 \times 300 = 10609 - 1200 = 9409 \). The discriminant is 9409.

Step 4: Solve for \( y \)

Substitute the discriminant back into the quadratic formula: \( y = \frac{103 \pm \sqrt{9409}}{2} \). Since \( \sqrt{9409} = 97 \), we have two solutions: \( y = \frac{103 + 97}{2} = 100 \) and \( y = \frac{103 - 97}{2} = 3 \).

Step 5: Solve for \( x \) in terms of \( y \)

Since \( y = 10^x \), we solve for \( x \) as follows: \( 10^x = 100 \Rightarrow x = \log_{10}(100) = 2 \) and \( 10^x = 3 \Rightarrow x = \log_{10}(3) \).

Step 6: Conclude Solutions

Thus, the solutions for \( x \) are \( x = 2 \) and \( x = \log_{10}(3) \).

Key Concepts

Factoring

When faced with a quadratic equation, one powerful technique is factoring. Factoring involves breaking down an expression into products of simpler expressions. In the context of quadratic equations, we aim to express the quadratic equation in the form

• \( ax^2 + bx + c = (px + q)(rx + s) \)

This approach involves determining two numbers that multiply to the constant term, \( c \), and add up to the linear coefficient, \( b \). For the equation derived from the exercise, \( y^2 - 103y + 300 = 0 \), we ultimately factor it into two binomials if possible. However, since others chose the quadratic formula method in the solution, factoring was considered bypassed here. If it had been possible to find such numbers, solutions could directly emerge from setting each binomial to zero. Factoring is most effective when dealing with smaller coefficients where the numbers involved are easily manageable.

Quadratic Formula

The quadratic formula is a fundamental method for solving any quadratic equation of the form \( ax^2 + bx + c = 0 \). The formula is expressed as follows:

• \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

This formula provides a direct way to find the roots of the equation, especially when equations are too complex for simple factoring. In the original problem, we recognized \( y = 10^x \) and then applied the quadratic formula using coefficients \( a = 1 \), \( b = -103 \), and \( c = 300 \). The essential part is accurately calculating the discriminant \( b^2 - 4ac \). If the discriminant is a perfect square, then the quadratic equation will have rational solutions, as in this case, where \( sqrt{9409} = 97 \). This method is incredibly versatile but requires care in execution to avoid errors with signs and arithmetic.

Exponential Equations

An exponential equation is one in which variables appear as exponents. The original problem involves such an equation because it includes the term \( 10^x \). When working with exponential equations, it's strategic to recognize patterns that allow transformations, such as expressing one part of the equation in the form of another. In our exercise, restructuring the equation led to substituting \( y = 10^x \), transforming the exponential equation into a quadratic one. This step simplifies solving, as calculus or logarithmic transformations might otherwise be needed. Thus, understanding how to manipulate exponential terms is key. In other scenarios, exponential equations might require logarithms or other advanced techniques, but here, the quadratic transformation proved efficient.

Logarithms

Logarithms are the inverse operations of exponential functions, used to solve expressions involving exponents. In the exercise, once we find the solutions for \( y \), we must revert \( y = 10^x \) to find \( x \). Employing logarithms provides the means: when \( 10^x = 100 \), it implies taking the base 10 logarithm of both sides:

• \( x = \log_{10}(100) = 2 \)
• For \( 10^x = 3 \): \( x = \log_{10}(3) \)

Logarithms simplify this transition from the base raised to a power back to just the power. They are crucial for solving exponential equations that lack simple numerical solutions. Familiarity with logarithmic properties, such as the power rule and product rule, will further empower handling more complex problems. Using logarithms converts otherwise complicated exponential scenarios into manageable arithmetic problems.