Verified Answer :

  • Use the product property of exponents to simplify the term inside the square root to \(x^{-1}\).
  • Apply the rule that the square root is equivalent to a fractional exponent of \(\frac{1}{2}\).
  • Use the power rule \((x^a)^b = x^{a \cdot b}\) to obtain the final simplified result.
  • Conclude that the equivalent expression is \(\boxed{x^{-\frac{1}{2}}}\).

Explanation

  1. Simplify the inside of the square root

The expression to simplify is \(\sqrt{x^5 \cdot x^{-6}}\). We first focus on the expression inside the square root, which is a product of two powers of the same base \(x\). Using the exponent property \(x^a \cdot x^b = x^{a+b}\), we can combine these powers:
\[x^5 \cdot x^{-6} = x^{5 + (-6)} = x^{-1}\]
Thus, the original expression simplifies to \(\sqrt{x^{-1}}\).

  1. Simplify the full expression using exponents

Next, we express the square root as a fractional exponent. The square root of a value \(u\) is equivalent to \(u^{\frac{1}{2}}\). Therefore:
\[\sqrt{x^{-1}} = (x^{-1})^{\frac{1}{2}}\]
We now apply the power of a power rule, which states \((x^a)^b = x^{a \cdot b}\). Multiplying the exponents, we have:
\[(x^{-1})^{\frac{1}{2}} = x^{-1 \cdot \frac{1}{2}} = x^{-\frac{1}{2}}\]
Thus, the expression is equivalent to \(x^{-\frac{1}{2}}\).

  1. Conclusion

Comparing our simplified result \(x^{-\frac{1}{2}}\) with the given options, we identify that the correct equivalent expression is \(x^{-\frac{1}{2}}\). Consequently, the final answer is \(\boxed{x^{-\frac{1}{2}}}\).

Examples

This calculation is a fundamental example of how exponents work in real-world scenarios. Imagine you are working with data compression where the size of a file is represented by a power of a variable \(x\). If you perform a series of transformations on the data that correspond to multiplying and taking roots of these power-based sizes, you need to understand how to combine the exponents to keep track of the final data size efficiently. Simplifying such expressions allows computer scientists to optimize algorithms and handle large-scale data processing effectively.