Question

question
fully simplify using only positive exponents.
\\(\frac{27xy^{7}}{x^{8}y^{2}}\\)

Explanation:

Step1: Simplify the coefficient and the \(x\)-terms

For the coefficient, it remains \(27\). For the \(x\)-terms, use the rule \(\frac{x^m}{x^n}=x^{m - n}\). Here, \(m = 1\) and \(n = 8\), so \(\frac{x}{x^8}=x^{1-8}=x^{-7}\). But we need positive exponents, so \(x^{-7}=\frac{1}{x^7}\), but actually, we can handle it as part of the fraction: \(\frac{x}{x^8}=\frac{1}{x^{8 - 1}}=\frac{1}{x^7}\) (wait, no, let's do it correctly. The rule is \(a^m\div a^n=a^{m - n}\), so \(\frac{x^1}{x^8}=x^{1-8}=x^{-7}=\frac{1}{x^7}\), but we can also think of it as moving \(x^1\) to the denominator as \(x^{-1}\) and then subtract? Wait, better to apply the exponent rule directly. So for the \(x\)-terms: \(\frac{x}{x^8}=x^{1 - 8}=x^{-7}\), and for the \(y\)-terms: \(\frac{y^7}{y^2}=y^{7 - 2}=y^5\). Then, \(x^{-7}=\frac{1}{x^7}\), but we can write the entire expression as \(27\times x^{1-8}\times y^{7 - 2}\).

Step2: Calculate the exponents

Calculate the exponents for \(x\) and \(y\). For \(x\): \(1-8=-7\), so \(x^{-7}\). For \(y\): \(7 - 2 = 5\), so \(y^5\). Now, rewrite \(x^{-7}\) as \(\frac{1}{x^7}\) to have positive exponents. So the expression becomes \(27\times\frac{1}{x^7}\times y^5=\frac{27y^5}{x^7}\). Wait, let's do it step by step:

First, separate the coefficient, \(x\)-terms, and \(y\)-terms:

\(\frac{27xy^7}{x^8y^2}=27\times\frac{x}{x^8}\times\frac{y^7}{y^2}\)

Now, apply the quotient rule for exponents (\(\frac{a^m}{a^n}=a^{m - n}\)):

For \(x\): \(\frac{x^1}{x^8}=x^{1 - 8}=x^{-7}\)

For \(y\): \(\frac{y^7}{y^2}=y^{7 - 2}=y^5\)

Now, rewrite \(x^{-7}\) using the negative exponent rule (\(a^{-n}=\frac{1}{a^n}\)):

\(x^{-7}=\frac{1}{x^7}\)

So now, substitute back:

\(27\times\frac{1}{x^7}\times y^5=\frac{27y^5}{x^7}\)

Wait, but let's check again. The coefficient is 27, \(x\) term: \(x^1\div x^8 = x^{1 - 8}=x^{-7}=\frac{1}{x^7}\), \(y\) term: \(y^7\div y^2 = y^{7 - 2}=y^5\). So multiplying them together: \(27\times\frac{1}{x^7}\times y^5=\frac{27y^5}{x^7}\).

Answer:

\(\dfrac{27y^{5}}{x^{7}}\)