Problem 43 - Precalculus: Functions and Graphs

Question

Find the real solutions of the equation. (a) \(x^{5 / 3}=32\) (b) \(x^{4 / 3}=16\) (c) \(x^{2 / 3}=-36\) (d) \(x^{34}=125\) (e) \(x^{3 / 2}=-27\)

Step-by-Step Solution

Answer

Real solutions: (a) 8, (b) 8; no real solutions for parts (c), (d), and (e).

Step 1: Understanding the Problem

You need to solve for the variable \(x\) in five different fractional exponent equations. Each equation asks for the real solutions only.

Step 2: Solving part (a)

Equation: \(x^{5/3} = 32\).To solve for \(x\), raise both sides to the power of \(3/5\):\[(x^{5/3})^{3/5} = 32^{3/5}\]Simplifying the left side gives \(x\):\[x = 32^{3/5}\]Calculate \(32^{3/5}\) by first evaluating \(32^{1/5} = 2\) and then \(2^3 = 8\). Thus, \(x = 8\).

Step 3: Solving part (b)

Equation: \(x^{4/3} = 16\).Raise both sides to the power of \(3/4\):\[(x^{4/3})^{3/4} = 16^{3/4}\]This simplifies to \(x = 16^{3/4}\).Evaluate \(16^{1/4} = 2\) and \(2^3 = 8\). So, \(x = 8\).

Step 4: Analyzing part (c)

Equation: \(x^{2/3} = -36\).Since there's no real number whose even root (like \(2/3\)) is negative, this equation has no real solution.

Step 5: Solving part (d)

Equation: \(x^{34} = 125\).Raise both sides to the power of \(1/34\):\[(x^{34})^{1/34} = 125^{1/34}\]Simplifying gives \(x = 125^{1/34}\). Since a precise computation of this is not straightforward by ordinary means, the real positive solution is considered to generally satisfy this form. A practical approximation or computational tool is typically needed to find explicit values for unconventional fractional powers.

Step 6: Analyzing part (e)

Equation: \(x^{3/2} = -27\).Since there's no real number whose square (part of an even root like \(3/2\)) results in a negative number, this equation has no real solution.

Key Concepts

Real Solutions

When we talk about finding real solutions in mathematics, we are specifically looking for values that satisfy a given equation without involving any imaginary numbers. Imaginary numbers, which include the square root of negative numbers, don't count as real solutions because they end up in a different number system altogether called the complex number system.

For any equation involving fractional exponents, the real solutions are those values of the variable that we can plug back into the equation to satisfy it entirely using just real numbers. For instance, in parts (a) and (b) of our exercise, we find real solutions by ensuring that when raised to a fractional power, the results are positive real numbers.

However, in parts (c) and (e), we encounter equations like \(x^{2/3} = -36\) and \(x^{3/2} = -27\). Here, the fractional exponents involve even roots, and since no real number exists whose even root yields a negative number, these equations do not have real solutions. Understanding this distinction is key to solving such problems in precalculus.

Solving Equations

Solving equations, especially those with fractional exponents, can initially seem challenging, but it becomes simpler once you grasp the basic rule of operations.

• The primary step is usually to isolate the term involving the variable.
  
• Next, you eliminate the fractional exponent by raising both sides of the equation to the reciprocal of that fraction. This is because \( (x^{a/b})^{b/a} = x\), thus solving for \(x\).
  
• Such a method ensures we see if a real solution exists and makes it easier to compute if it does.
  

For instance, in part (a), the equation \(x^{5/3} = 32\) transforms into \((x^{5/3})^{3/5} = 32^{3/5}\). Simplifying the left side leaves us with the variable \(x\), and provides a neat path to the solution. This method checks if the original equation can result in a real number solution.

However, it's important to remember that not all equations allow real solutions. Equations like \(x^{3/2} = -27\) prove unsolvable in real terms due to the negative result of the operation, indicating the importance of identifying equation limitations at an early stage.

Precalculus

Precalculus is a foundational mathematical course bridging algebra and calculus. It prepares students for the concepts and challenges of calculus. Fractional exponents are a critical part of precalculus because they extend the concept of root operations, such as square roots and cube roots, to include other fractional values.

Learning to handle fractional exponents in precalculus is crucial since it equips students with the knowledge of simplifying complex expressions and solving intricate equations. For example, understanding how to convert radical expressions into fractional exponents can simplify complex algebraic problems.

• By understanding the mechanics behind expressions like \(32^{3/5}\), students build a robust mathematical toolkit for future calculus problems.
  
• This understanding also enhances mathematical reasoning skills, allowing learners to analyze patterns and anticipate outcomes more effectively.
  

Precalculus thus not only serves as a preparation for calculus but also strengthens students' algebraic skills, enabling them to approach mathematical challenges with enhanced confidence and skills.