Problem 92 - College Algebra Essentials

Question

Find all values of \(x\) satisfying the given conditions. $$y=(x-5)^{\frac{3}{2}} \text { and } y=125$$

Step-by-Step Solution

Answer

The value of \(x\) that satisfies the conditions is \(x = 10\).

Step 1: Substitute values

In the equation \(y = (x - 5)^{3/2}\), replace \(y\) with the given value of 125: \(125 = (x - 5)^{3/2}\).

Step 2: Obtain square root

To isolate the term containing x, we will need to square root both sides of the equation. When we do this, we have \(\sqrt[3]{125} = x - 5\).

Step 3: Isolate x

Finally, solve for x by adding 5 to both sides of the equation. This gives \(x = \sqrt[3]{125} + 5\).

Key Concepts

Exponents and Radicals

When we work with exponents and radicals, we are dealing with two sides of the same coin. An exponent, like the one found in \( (x - 5)^{\frac{3}{2}} \), represents repeated multiplication. For instance, \( a^2 \) means \( a \) multiplied by itself. On the other hand, a radical is an expression that involves roots, such as square roots \( \sqrt{a} \) or cube roots \( \sqrt[3]{a} \). The cube root, which you’re likely to encounter in our original problem, is the inverse operation of raising a number to the exponent of 3. Hence, if \( b^3 = a \), then we can say \( b = \sqrt[3]{a} \).

In our textbook solution, we used the fact that raising a number to the exponent of \( \frac{3}{2} \) is equivalent to taking the square root of that number after it has been cubed. This is vital to grasp because it allows you to reverse the operation. By understanding how exponents relate to multiplication and how radicals express the concept of roots, you unlock the ability to manipulate and solve various algebraic equations involving these terms.

Isolate Variable

Isolating the variable, often the goal in algebra, refers to the process of manipulating an equation to get the variable by itself on one side of the equals sign. It's a fundamental skill for solving equations and understanding how different operations can be reversed or undone to reveal the value of the variable in question. Consider our original exercise where we have \( (x - 5)^{\frac{3}{2}} = 125 \).

To isolate \( x \) in this scenario, we performed operations that undo the exponent on the left side. By taking the cube root of both sides, we reverse the effect of raising \( x - 5 \) to the power of 3/2. Once we have removed the radical, we're left with \( x - 5 = \sqrt[3]{125} \), a much simpler equation. Adding 5 to both sides then fully isolates \( x \), allowing us to find its value. The ability to isolate variables is essential to solve virtually any algebraic equation you'll come across.

Cube Roots

The cube root of a number is a special value that, when used as a factor three times, gives the original number. The cube root of \( a \) is written as \( \sqrt[3]{a} \). For example, the cube root of 8 is 2, because \( 2 \times 2 \times 2 = 8 \). Cube roots can be applied to both positive and negative numbers, unlike square roots which are only defined for non-negative numbers.

In the context of our exercise, we come across \( \sqrt[3]{125} \), which is the cube root of 125. Since \( 5 \times 5 \times 5 = 125 \), it follows that the cube root of 125 is 5. Recognizing this relationship allows us to conclude that \( \sqrt[3]{125} = 5 \), thus simplifying our equation from Step 2 of the solution to \( x = 5 + 5 \). Understanding cube roots is particularly important when you're dealing with equations where the variable is raised to the power of 3, or when you're working with volumes of cubic shapes where dimensions are equal.