Problem 46 - Elementary and Intermediate Algebra

Question

Solve each equation. Let \(f(x)=\sqrt[5]{4 x-4} .\) For what value(s) of \(x\) is \(f(x)=-2 ?\)

Step-by-Step Solution

Answer

The value of \(x\) is -7.

Step 1: Understand the Function

The given function is \(f(x) = \sqrt[5]{4x - 4}\). This means we need to find the value of \(x\) for which \(f(x) = -2\). In other words, \(\sqrt[5]{4x - 4} = -2\).

Step 2: Set the Equation Equal to -2

Set the function \(f(x)\) equal to -2: \(\sqrt[5]{4x - 4} = -2\). Our goal now is to solve for \(x\).

Step 3: Remove the Fifth Root

To eliminate the fifth root, raise both sides of the equation to the fifth power:\[(\sqrt[5]{4x - 4})^5 = (-2)^5\]This simplifies to: \[4x - 4 = -32\].

Step 4: Solve for x

Now solve the linear equation \(4x - 4 = -32\). Start by adding 4 to both sides:\[4x = -28\]Then, divide both sides by 4:\[x = -7\]

Step 5: Verify the Solution

Substitute \(x = -7\) back into the original function to verify the solution. Calculate:\[f(-7) = \sqrt[5]{4(-7) - 4} = \sqrt[5]{-28 - 4} = \sqrt[5]{-32}\]Since \((-32)^{1/5} = -2\), \(f(-7) = -2\). This confirms \(x = -7\) is the correct solution.

Key Concepts

Fifth Roots

Fifth roots can be a bit tricky at first, but they're merely the inverse operation of raising a number to the fifth power. Simply put, if you have a number and want to find its fifth root, you're figuring out what number needs to be multiplied by itself five times to give you the original number. For instance, the fifth root of \(-32\) can be found by asking, "What number to the power of five equals \(-32\)?" The answer is \(-2\), because when you multiply \(-2\) by itself five times, you get \(-32\).

• When "raising" a number to a fraction like \(1/5\), you're essentially finding the fifth root.
• The process involves reversing the fifth power operation—think of it as "unpacking" the number down to its root!
• Fifth roots can be both positive and negative, depending on the original number; for negative numbers, the root is negative to maintain the sign.

To solve equations involving fifth roots, you typically isolate the root, then "raise" both sides of the equation to the power of five to eliminate the root and solve for the variable.

Functions in Algebra

Functions are like little machines in algebra that take an input and give an output. Think of them as similar to a vending machine: you input a coin, make a selection (inputting), and out comes your snack (output). In the algebraic world, a function usually tells you how those inputs and outputs relate through an equation.

• The function \(f(x) = \sqrt[5]{4x - 4}\), for example, takes an \(x\) and outputs the fifth root of \(4x - 4\).
• Often, you'll be asked to "solve" a function by determining which \(x\) values lead to specific outputs. This means working backwards from a given output to find the corresponding input.
• In the case above, we sought an \(x\) such that the function output is \(-2\). This meant "reversing" the function process to find the input.

Understanding functions in algebra is crucial since they allow us to describe and analyze relationships between variables.

Linear Equations

Linear equations crop up all the time in algebra. They're the building blocks of many more complex algebraic concepts. A linear equation looks like a straight line when plotted on a graph, because it maintains a constant rate of change. At its core, solving a linear equation involves finding the value of \(x\) that makes the equation true.

• The linear equation from our exercise, \(4x - 4 = -32\), forms after removing the fifth root.
• Solving these equations usually means performing operations like addition, subtraction, multiplication, or division to isolate \(x\).
• In today's example, once we isolate \(x\), we see the steps: first add 4 to both sides, then divide by 4, efficiently solving for \(x\).

Recognizing and working through linear equations is an essential skill in algebra, preparing you for more advanced topics in mathematics. Dexterity with them will serve you well across various math problems, especially when functions and roots step onto the stage.