Problem 24 - Algebra 2 and Trigonometry

Question

Solve for \(x\) and check: \(\frac{x^{\frac{1}{3}}}{x^{\frac{2}{3}}}=10 .\) Use the rule for the division of powers with like bases to simplify the left side of the equation.

Step-by-Step Solution

Answer

The solution is \( x = \frac{1}{1000} \).

Step 1: Apply the Division Rule for Exponents

Begin by applying the division rule for exponents, which states that when dividing like bases, you subtract the exponents. Here, the expression is \( \frac{x^{\frac{1}{3}}}{x^{\frac{2}{3}}} \). Subtract the exponents: \( \frac{1}{3} - \frac{2}{3} = -\frac{1}{3} \). This gives: \( x^{-\frac{1}{3}} \).

Step 2: Simplify the Equation

Now that you have \( x^{-\frac{1}{3}} \), you set the equation equal to 10: \( x^{-\frac{1}{3}} = 10 \).

Step 3: Solve for x

To eliminate the negative exponent, rewrite the equation in terms of positive exponents: \( \frac{1}{x^{\frac{1}{3}}} = 10 \). Then, take the reciprocal of both sides to get: \( x^{\frac{1}{3}} = \frac{1}{10} \).

Step 4: Cube Both Sides

To find \(x\), cube both sides of the equation to get rid of the fractional exponent: \( (x^{\frac{1}{3}})^3 = \left(\frac{1}{10}\right)^3 \). This simplifies to \( x = \frac{1}{1000} \).

Step 5: Check the Solution

Substitute \( x = \frac{1}{1000} \) back into the original equation: The left side becomes \( \frac{\left(\frac{1}{1000}\right)^{\frac{1}{3}}}{\left(\frac{1}{1000}\right)^{\frac{2}{3}}} \), which, by applying the division rule for exponents, simplifies to \( \left(\frac{1}{1000}\right)^{-\frac{1}{3}} \). Simplifying further gives 10, confirming that the solution is correct.

Key Concepts

Solving Equations

When working with equations, our goal is to isolate the variable, typically represented by a letter such as \(x\). This means finding the value of \(x\) that makes the equation true.
Let's revisit how we solved the given equation step by step. First, we had an equation involving division of powers: \(\frac{x^{\frac{1}{3}}}{x^{\frac{2}{3}}} = 10\).
Using the division rule for exponents, we simplified the left side by subtracting the exponents, resulting in \(x^{-\frac{1}{3}} = 10\). Now, the equation becomes simpler.
The challenge was to eliminate the negative exponent to solve for \(x\). By rewriting \(x^{-\frac{1}{3}}\) as \(\frac{1}{x^{\frac{1}{3}}}\), we further simplified the equation.
Then, taking the reciprocal of both sides gave us \(x^{\frac{1}{3}} = \frac{1}{10}\).
To solve for \(x\), we cubed both sides to eliminate the fractional exponent. This led us to \(x = \frac{1}{1000}\). Thus, solving equations often involves several algebraic manipulations to isolate the variable.

Negative Exponents

Negative exponents can often seem tricky, but they are simply another way to express division. When you see \(x^{-a}\), it can be rewritten as \(\frac{1}{x^a}\). This means you're dividing 1 by \(x\) raised to the power of \(a\).
In our exercise, once we simplified the exponents in the equation, \(x^{-\frac{1}{3}}\) indicated a need to express the exponent positively.
That transformation resulted in \(\frac{1}{x^{\frac{1}{3}}}\).
By rewriting and understanding negative exponents, algebra problems become much more manageable.
So remember, whenever you encounter a negative exponent, think of it as instructing you to take the reciprocal.

Fractional Exponents

Fractional exponents can be thought of as a way to express roots. When you have an exponent like \(x^{\frac{1}{3}}\), it means you're finding the cube root of \(x\).
In our solution, we encountered \(x^{\frac{1}{3}}\) and needed to eliminate the fractional part to solve for \(x\).
The way we handled this was by raising both sides of the equation to the power of 3—since cubing is the inverse operation of taking the cube root.
This transformed \((x^{\frac{1}{3}})^3\) into just \(x\), and it turned \(\left(\frac{1}{10}\right)^3\) into \(\frac{1}{1000}\).
Understanding fractional exponents is crucial because they frequently appear in various equations. They allow us to transition effortlessly between roots and powers, creating flexibility in solving complex equations.