Question

1. fill in the blanks to make true equations.

27 -\frac{1}{2} \frac{2}{3} y \frac{7}{2}
81 -4 6 x 3
\left(x \frac{1}{3} y^{\square}\
ight)^{3}=x y^{2} \left(x^{4} y^{-2}\
ight)^{\square}=\frac{y}{x^{2}} \left(64 x^{-7} y^{2}\
ight)^{\frac{1}{2}}=\frac{8 y}{x^{\square}}
\left(x^{\frac{1}{5}} y^{3}\
ight)\left(x^{\frac{2}{5}} y^{\square}\
ight)=\frac{x^{\frac{3}{5}}}{y} \frac{\left(\square x^{3}\
ight)^{\frac{2}{3}}}{3 x^{5}}=\frac{3}{x^{3}} \left(\frac{x^{\frac{1}{2}}}{y^{\frac{1}{3}}}\
ight)^{-6}=\frac{\square^{2}}{\square^{3}}

1. reinforce under what conditions will a simplified expression still have a rational exponent?

Explanation:

First Equation: \((x^{\frac{1}{3}}y^{\square})^3 = xy^2\)

Step1: Apply the power - of - a - product rule \((ab)^n=a^nb^n\)

\((x^{\frac{1}{3}}y^{\square})^3=x^{\frac{1}{3}\times3}y^{3\times\square}\)
\(=x^1y^{3\times\square}\)

Step2: Equate the exponents of \(y\)

Since \(x^1y^{3\times\square}=xy^2\), then \(3\times\square = 2\). Solving for \(\square\), we get \(\square=\frac{2}{3}\)

Second Equation: \((x^{4}y^{- 2})^{\square}=\frac{y}{x^{2}}\)

Step1: Apply the power - of - a - product rule \((ab)^n=a^nb^n\)

\((x^{4}y^{-2})^{\square}=x^{4\square}y^{-2\square}\)

Step2: Equate the exponents of \(x\) and \(y\)

For \(x\): \(4\square=- 2\), so \(\square =-\frac{1}{2}\). For \(y\): \(-2\square = 1\), substituting \(\square =-\frac{1}{2}\), \(-2\times(-\frac{1}{2})=1\)

Third Equation: \((64x^{-7}y^{2})^{\frac{1}{2}}=\frac{8y}{x^{\square}}\)

Step1: Apply the power - of - a - product rule \((ab)^n=a^nb^n\)

\((64x^{-7}y^{2})^{\frac{1}{2}}=64^{\frac{1}{2}}x^{-7\times\frac{1}{2}}y^{2\times\frac{1}{2}}\)
\(=8x^{-\frac{7}{2}}y\)

Step2: Rewrite in the form \(\frac{8y}{x^{\square}}\)

Since \(8x^{-\frac{7}{2}}y=\frac{8y}{x^{\frac{7}{2}}}\), then \(\square=\frac{7}{2}\)

Fourth Equation: \((x^{\frac{1}{5}}y^{3})(x^{\frac{2}{5}}y^{\square})=\frac{x^{\frac{3}{5}}}{y}\)

Step1: Apply the product rule \(a^m\times a^n=a^{m + n}\)

\((x^{\frac{1}{5}}y^{3})(x^{\frac{2}{5}}y^{\square})=x^{\frac{1 + 2}{5}}y^{3+\square}\)
\(=x^{\frac{3}{5}}y^{3+\square}\)

Step2: Equate the exponents of \(y\)

Since \(x^{\frac{3}{5}}y^{3+\square}=\frac{x^{\frac{3}{5}}}{y}=x^{\frac{3}{5}}y^{-1}\), then \(3+\square=-1\). Solving for \(\square\), we get \(\square=-4\)

Fifth Equation: \(\frac{(\square x^{3})^{\frac{2}{3}}}{3x^{5}}=\frac{3}{x^{3}}\)

Step1: Apply the power - of - a - product rule \((ab)^n=a^nb^n\)

Let the unknown be \(a\). \(\frac{(ax^{3})^{\frac{2}{3}}}{3x^{5}}=\frac{a^{\frac{2}{3}}x^{2}}{3x^{5}}\)
\(=\frac{a^{\frac{2}{3}}}{3x^{3}}\)

Step2: Equate the expressions

Since \(\frac{a^{\frac{2}{3}}}{3x^{3}}=\frac{3}{x^{3}}\), then \(a^{\frac{2}{3}} = 9\). Raising both sides to the \(\frac{3}{2}\) power, \(a = 9^{\frac{3}{2}}=(3^2)^{\frac{3}{2}}=27\)

Sixth Equation: \((\frac{x^{\frac{1}{2}}}{y^{\frac{1}{3}}})^{-6}=\frac{\square^{2}}{\square^{3}}\)

Step1: Apply the power - of - a - quotient rule \((\frac{a}{b})^n=\frac{a^n}{b^n}\)

\((\frac{x^{\frac{1}{2}}}{y^{\frac{1}{3}}})^{-6}=\frac{x^{-3}}{y^{-2}}\)
\(=\frac{y^{2}}{x^{3}}\)

Answer:

First blank: \(\frac{2}{3}\), Second blank: \(-\frac{1}{2}\), Third blank: \(\frac{7}{2}\), Fourth blank: \(-4\), Fifth blank: \(27\), Sixth blank: \(y\) and \(x\) (in the fraction \(\frac{y^{2}}{x^{3}}\))