Which expression is equivalent to $\frac{b^2}{a b^{-3}}$? Assume $a
eq 0, b
eq 0$.


A$\frac{a}{b^5}$
B$\frac{1}{a b^5}$
C$\frac{a^3 b}{1}$
D$\frac{b}{a}$

Verified Answer :

  • Identify the expression: \(\frac{b^2}{a b^{-3}}\).
  • Apply the negative exponent rule: \(\frac{1}{b^{-3}} = b^3\).
  • Simplify the product: \(b^2 \cdot b^3 = b^{2+3} = b^5\).
  • Final simplified expression: \(\frac{b^5}{a}\).
  • The calculated result is \(\boxed{\frac{b^5}{a}}\).

Explanation

  1. Analyzing the Expression and Applying Exponent Rules

The expression to simplify is \(\frac{b^2}{a b^{-3}}\), under the conditions that \(a eq 0\) and \(b eq 0\). To simplify this, we use the laws of exponents. Specifically, the negative exponent rule states that \(\frac{1}{b^{-n}} = b^n\). Applying this, we can rewrite the term \(b^{-3}\) in the denominator by moving it to the numerator, which changes its exponent to positive. This yields \(\frac{1}{b^{-3}} = b^3\). Therefore, the expression becomes \(\frac{b^2 \cdot b^3}{a}\).

  1. Simplifying the Numerator

Next, we use the product rule for exponents, which states that \(b^m \cdot b^n = b^{m+n}\). Applying this to the numerator, we have \(b^2 \cdot b^3 = b^{2+3} = b^5\). Consequently, the entire expression simplifies to \(\frac{b^5}{a}\).

  1. Conclusion and Evaluation of Options

Upon evaluating the given options (\(\frac{a}{b^5}\), \(\frac{1}{a b^5}\), \(\frac{a^3 b}{1}\), and \(\frac{b}{a}\)), it appears that none of these choices are mathematically equivalent to the simplified expression \(\frac{b^5}{a}\). It is likely that there is a typographical error in the provided options or the original expression. However, following standard algebraic principles, the expression \(\frac{b^2}{a b^{-3}}\) simplifies strictly to \(\frac{b^5}{a}\).

Examples

Simplifying expressions with exponents is a fundamental skill in algebra, commonly used in fields such as engineering, physics, and computer science. For example, when calculating electrical resistance or signal decay, you might encounter complex formulas like \(\frac{R_1 \cdot I^{-2}}{R_2 \cdot I^3}\). Simplifying these helps in understanding how variables like current (\(I\)) and resistance (\(R\)) interact, allowing for more efficient problem-solving and system optimization. Mastering these algebraic manipulations ensures you can handle more advanced topics with confidence.