Question

question
select the equivalent expression.
\\(\frac{x^{-8}y^{-3}y}{x^{-5}}\\)
answer
\\(\circ\\) \\(\frac{x^3}{y^4}\\)
\\(\circ\\) \\(\frac{1}{x^3y^2}\\)
\\(\circ\\) \\(\frac{y^4}{x^3}\\)
\\(\circ\\) \\(x^3y^2\\)

Explanation:

Step1: Simplify y terms

For the y - terms in the numerator, we have \(y^{-3} \cdot y\). Using the exponent rule \(a^m \cdot a^n=a^{m + n}\), here \(a = y\), \(m=-3\) and \(n = 1\), so \(y^{-3}\cdot y=y^{-3 + 1}=y^{-2}\). Now the expression becomes \(\frac{x^{-8}y^{-2}}{x^{-5}}\).

Step2: Simplify x terms

Using the exponent rule \(\frac{a^m}{a^n}=a^{m - n}\) for the x - terms, where \(a=x\), \(m=-8\) and \(n=-5\). So \(x^{-8-(-5)}=x^{-8 + 5}=x^{-3}\). Now the expression is \(\frac{x^{-3}y^{-2}}{1}\) (since the denominator after simplifying x terms is 1).

Step3: Rewrite with positive exponents

Recall that \(a^{-n}=\frac{1}{a^{n}}\). So \(x^{-3}=\frac{1}{x^{3}}\) and \(y^{-2}=\frac{1}{y^{2}}\). Then \(x^{-3}y^{-2}=\frac{1}{x^{3}}\cdot\frac{1}{y^{2}}=\frac{1}{x^{3}y^{2}}\).

Answer:

\(\frac{1}{x^{3}y^{2}}\) (the second option from the left, i.e., the option \(\boldsymbol{\frac{1}{x^{3}y^{2}}}\))