- Apply the rule \(\left(\frac{a}{b}\right)^{-n} = \left(\frac{b}{a}\right)^n\) to invert the fraction: \(\left(\frac{3 y^{-6}}{2 x^{-5}}\right)^{3}\).
- Distribute the power of 3 to each term: \(\frac{3^3 \cdot (y^{-6})^3}{2^3 \cdot (x^{-5})^3} = \frac{27 y^{-18}}{8 x^{-15}}\).
- Use the rule \(a^{-n} = \frac{1}{a^n}\) to rewrite with positive exponents: \(\frac{27 x^{15}}{8 y^{18}}\).
- Final simplified result: \(\boxed{\frac{27 x^{15}}{8 y^{18}}}\).
Explanation
- Analyzing the Problem
We are given the expression \(\left(\frac{2 x^{-5}}{3 y^{-6}}\right)^{-3}\) and tasked with simplifying it such that all exponents are positive. We assume \(x, y
eq 0\). The expression involves properties of exponents, specifically the power of a quotient and the power of a power rule. The core rules are:
- \(\left(\frac{a}{b}\right)^{-n} = \left(\frac{b}{a}\right)^n\)
- \((a \cdot b)^n = a^n \cdot b^n\)
- \(\left(a^m\right)^n = a^{mn}\)
- \(a^{-n} = \frac{1}{a^n}\)
- Inverting the Fraction
To handle the negative exponent outside the fraction, we use the rule \(\left(\frac{a}{b}\right)^{-n} = \left(\frac{b}{a}\right)^n\).
Applying this to our expression:
\[\left(\frac{2 x^{-5}}{3 y^{-6}}\right)^{-3} = \left(\frac{3 y^{-6}}{2 x^{-5}}\right)^{3}\]
- Distributing the Exponent
Now, apply the exponent of 3 to each factor in the numerator and the denominator:
\[\left(\frac{3 y^{-6}}{2 x^{-5}}\right)^{3} = \frac{3^3 \cdot (y^{-6})^3}{2^3 \cdot (x^{-5})^3}\]
Calculating the numerical coefficients:
\(3^3 = 27\)
\(2^3 = 8\)
Using the rule \(\left(a^m\right)^n = a^{mn}\) for the variables:
\((y^{-6})^3 = y^{-6 \cdot 3} = y^{-18}\)
\((x^{-5})^3 = x^{-5 \cdot 3} = x^{-15}\)
Combining these:
\[\frac{27 \cdot y^{-18}}{8 \cdot x^{-15}}\]
- Ensuring Positive Exponents
To express the result with positive exponents, we use the rule \(a^{-n} = \frac{1}{a^n}\). This means a term with a negative exponent in the numerator moves to the denominator as a positive exponent, and vice versa:
- \(y^{-18}\) in the numerator becomes \(y^{18}\) in the denominator.
- \(x^{-15}\) in the denominator becomes \(x^{15}\) in the numerator.
The expression becomes:
\[\frac{27 \cdot x^{15}}{8 \cdot y^{18}}\]
All exponents are now positive.
Examples
In science and engineering, we often deal with extremely large or small quantities expressed using scientific notation or power laws. For example, if you are calculating the volume of a very small scale structure where dimensions involve negative powers of a base unit, the ability to simplify these expressions with positive exponents makes the physical interpretation and subsequent calculations much easier to handle without errors. This algebraic technique is fundamental for solving physics and chemistry problems efficiently.