Problem 45 - Precalculus with Calculus Previews

Question

Either use factoring or the quadratic formula to solve the given equation. $$ \left(5^{x}\right)^{2}-2\left(5^{x}\right)-1=0 $$

Step-by-Step Solution

Answer

The solution for \( x \) is \( x = \frac{\log(1 + \sqrt{2})}{\log(5)} \).

Step 1: Identify the Substitution

Recognize that the equation can be written in terms of a single variable by letting \( y = 5^x \). Rewrite the equation as \( y^2 - 2y - 1 = 0 \).

Step 2: Determine the Discriminant

In a quadratic equation in the form \( ay^2 + by + c = 0 \), calculate the discriminant using \( b^2 - 4ac \). For our equation, \( a = 1 \), \( b = -2 \), and \( c = -1 \). Thus, the discriminant is \((-2)^2 - 4(1)(-1) = 4 + 4 = 8\).

Step 3: Solve Using the Quadratic Formula

Apply the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Substitute the values from our equation to get \( y = \frac{2 \pm \sqrt{8}}{2} \). Simplifying \( \sqrt{8} = 2\sqrt{2} \), the solutions become \( y = \frac{2 \pm 2\sqrt{2}}{2} = 1 \pm \sqrt{2} \).

Step 4: Inverse Substitution to Solve for x

Use the solutions for \( y \) to find \( x \). We have \( 5^x = 1 + \sqrt{2} \) and \( 5^x = 1 - \sqrt{2} \). Since \( 1 - \sqrt{2} \) is negative and powers of 5 are positive, it is not possible. Thus, we have \( 5^x = 1 + \sqrt{2} \).

Step 5: Solve the Remaining Equation

Solve \( 5^x = 1 + \sqrt{2} \) by taking the logarithm on both sides: \( x \log(5) = \log(1 + \sqrt{2}) \). Therefore, \( x = \frac{\log(1 + \sqrt{2})}{\log(5)} \).

Key Concepts

Factoring

Factoring is a mathematical process used to break down complex expressions into simpler ones, often making equations easier to solve. It involves finding two or more expressions whose product equals the original expression. In the case of quadratic equations, like the one you've encountered, factoring helps express the quadratic in a product of binomials.

• Consider the quadratic equation in standard form: \( ax^2 + bx + c = 0 \).
• To factor, look for two numbers that multiply to \( ac \) and add up to \( b \).
• If successful, the quadratic breaks down into \((dx + e)(fx + g) = 0\), where \(d = a\) and \(f = 1\).

In our problem, the original equation was transformed using substitution into \( y^2 - 2y - 1 = 0 \). Sometimes equations aren't easily factored by conventional means, and the quadratic formula or another method may be required. When factoring doesn’t directly apply, it’s helpful to have a mixed approach, prepare to switch methods if one is not feasible. Remember, the ability to factor depends on the nature of the equation and the integers involved.

Quadratic Formula

The quadratic formula is a universal tool for solving any quadratic equation of the form \( ax^2 + bx + c = 0 \). It is especially useful when the equation is not easily factored. The formula is:\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

• The "\( b^2 - 4ac \)" part is known as the discriminant.
• The square root symbol "\( \pm \)" indicates there may be two solutions.
• Depending on the discriminant’s value, you can have two real solutions, one real solution, or two complex ones.

In the exercise, the discriminant was positive (\(8\)), indicating two real solutions: \(y = 1 + \sqrt{2}\) and \(y = 1 - \sqrt{2}\). This formula helps in cases where factoring might not be straightforward, making it a reliable alternative.

Logarithms

Logarithms are the inverse operations to exponentiation, used to solve equations where the unknown variable appears as an exponent. They can be particularly helpful when dealing with exponential equations such as \(5^x = y\). Here's how they work:

• The logarithm \( \log_b(a) \) answers the question: "To what power must \( b \) be raised, to get \( a \)?"
• Common logarithms use base 10, while natural logarithms use base \( e \).
• Logarithms convert multiplicative relationships to additive, making calculations more manageable.

In solving \(5^x = 1 + \sqrt{2}\), we took logarithms on both sides to isolate \(x\): \[ x \log(5) = \log(1 + \sqrt{2}) \] Thus, \[ x = \frac{\log(1 + \sqrt{2})}{\log(5)} \]. This technique simplifies the problem by transforming it into a more straightforward arithmetic task.