To express each of the given expressions with a single rational exponent, we will follow these steps:
1. Fourth root of \(x^3\)
The fourth root of a number can be expressed as that number raised to the power of \(1/4\). Thus, the fourth root of \(x^3\) can be expressed as: \[ \sqrt[4]{x^3} = (x^3)^{1/4} \] Using the power of a power rule \((a^m)^n = a^{m \cdot n}\), we get: \[ (x^3)^{1/4} = x^{3 \cdot \frac{1}{4}} = x^{\frac{3}{4}} \]
2. One over \(x^{-1}\)
This expression can be rewritten using the property of negative exponents, which states that \(a^{-n} = \frac{1}{a^n}\): \[ \frac{1}{x^{-1}} = x^{1} \]
3. Tenth root of the quantity \(x^5 \cdot x^4 \cdot x^2 \)
First, simplify the expression inside the parentheses. We can combine the exponents by adding them (since \(x^m \cdot x^n = x^{m+n}\)): \[ x^5 \cdot x^4 \cdot x^2 = x^{5 + 4 + 2} = x^{11} \] Now, we express the tenth root: \[ \sqrt[10]{x^{11}} = (x^{11})^{1/10} \] Using the power of a power rule again: \[ (x^{11})^{1/10} = x^{\frac{11}{10}} \]
4. \(x^{1/3} \cdot x^{1/3} \cdot x^{1/3}\)
Use the property of exponents (as shown previously) to combine the terms: \[ x^{\frac{1}{3}} \cdot x^{\frac{1}{3}} \cdot x^{\frac{1}{3}} = x^{\frac{1}{3} + \frac{1}{3} + \frac{1}{3}} = x^{\frac{1}{3} \cdot 3} = x^{1} \]
Summary of Expressions with Single Rational Exponents:
-
Fourth root of \(x^3\) can be rewritten as:
\[x^{\frac{3}{4}}\] -
One over \(x^{-1}\) can be rewritten as:
\[x^{1}\] -
Tenth root of \(x^5 \cdot x^4 \cdot x^2\) can be rewritten as:
\[x^{\frac{11}{10}}\] -
\(x^{1/3} \cdot x^{1/3} \cdot x^{1/3}\) can be rewritten as:
\[x^{1}\]
Which Expressions are Equivalent?
Now we can identify which expressions are equivalent based on their final forms:
- The expression \(x^{1}\) appears from both the second expression (\(1/x^{-1}\)) and the fourth expression (\(x^{1/3} \cdot x^{1/3} \cdot x^{1/3}\)).
- The expressions \(x^{\frac{3}{4}}\) and \(x^{\frac{11}{10}}\) are different from each other and from \(x^1\).
Thus, the equivalent expressions are:
- One over \(x^{-1}\) and \(x^{1/3} \cdot x^{1/3} \cdot x^{1/3}\) because both simplify to \(x\).