Question

exponent rules
simplify. your answer shou

1. \\(\frac{x^2y^2}{3y^3}\\)
2. \\(\frac{2x^0y^{-3}}{4xy^2}\\)
3. \\(\frac{4ba^2 \cdot 3ab}{3a^{-3}b^{-4}}\\)

Explanation:

Problem 1: Simplify \(\boldsymbol{\frac{x^2y^2}{3y^3}}\)

Step1: Apply exponent rule for \(y\)

When dividing exponents with the same base, subtract the exponents: \(y^a\div y^b = y^{a - b}\). Here, \(a = 2\), \(b = 3\), so \(y^{2-3}=y^{-1}=\frac{1}{y}\).

Step2: Simplify the expression

The \(x^2\) remains as is, and the coefficient \(3\) remains. So we have \(\frac{x^2}{3y}\).

Step1: Simplify \(x^0\)

Any non - zero number to the power of \(0\) is \(1\), so \(x^0 = 1\). The expression becomes \(\frac{2\times1\times y^{-3}}{4xy^2}=\frac{2y^{-3}}{4xy^2}\).

Step2: Simplify the coefficient

Simplify \(\frac{2}{4}=\frac{1}{2}\).

Step3: Apply exponent rules for \(x\) and \(y\)

For \(x\): \(x^{0 - 1}=x^{-1}=\frac{1}{x}\) (using \(x^a\div x^b=x^{a - b}\), here \(a = 0\), \(b = 1\)). For \(y\): \(y^{-3-2}=y^{-5}=\frac{1}{y^5}\) (using \(y^a\div y^b = y^{a - b}\), here \(a=-3\), \(b = 2\)).

Step4: Combine the terms

Multiply the simplified coefficient, \(x\) - term, and \(y\) - term: \(\frac{1}{2}\times\frac{1}{x}\times\frac{1}{y^5}=\frac{1}{2xy^5}\).

Step1: Simplify the numerator

Multiply the coefficients and apply the exponent rule for multiplication (\(a^m\times a^n=a^{m + n}\)) for \(a\) and \(b\) in the numerator.
Coefficients: \(4\times3 = 12\). For \(a\): \(a^{2+1}=a^3\). For \(b\): \(b^{1 + 1}=b^2\). So the numerator is \(12a^3b^2\).

Step2: Apply exponent rules for division

For the coefficient: \(\frac{12}{3}=4\). For \(a\): \(a^{3-(-3)}=a^{3 + 3}=a^6\) (using \(a^m\div a^n=a^{m - n}\), here \(m = 3\), \(n=-3\)). For \(b\): \(b^{2-(-4)}=b^{2 + 4}=b^6\) (using \(b^m\div b^n=b^{m - n}\), here \(m = 2\), \(n=-4\)).

Step3: Combine the terms

Multiply the coefficient, \(a\) - term, and \(b\) - term: \(4a^6b^6\).

Answer:

\(\frac{x^2}{3y}\)

Problem 3: Simplify \(\boldsymbol{\frac{2x^0y^{-3}}{4xy^2}}\)