Question

station 2
use the laws of exponents to simplify. write your
answers in positive exponents.

1. $(2x^{-2}y^{3})^{2}(6x^{-3}y^{-4})^{-2}$

1. $\frac{(a^{2}b^{4})^{\frac{3}{2}}}{(a^{-2}b^{-\frac{1}{2}})^{\frac{1}{3}}}$

1. $(\frac{1}{2}m^{-2}n^{\frac{1}{2}}p^{4})^{-2}(m^{2}n^{\frac{3}{2}}p^{0})^{2}$

Explanation:

Problem 3:

Step 1: Apply power of a product rule

For \((2x^{-2}y^{3})^{2}\), we have \(2^{2}(x^{-2})^{2}(y^{3})^{2}=4x^{-4}y^{6}\).
For \((6x^{-3}y^{-4})^{-2}\), we have \(6^{-2}(x^{-3})^{-2}(y^{-4})^{-2}=\frac{1}{36}x^{6}y^{8}\).

Step 2: Multiply the two results

Multiply \(4x^{-4}y^{6}\) and \(\frac{1}{36}x^{6}y^{8}\):
\(4\times\frac{1}{36} \cdot x^{-4 + 6} \cdot y^{6+8}=\frac{1}{9}x^{2}y^{14}\).

Problem 4:

Step 1: Simplify numerator and denominator using power of a power rule

Numerator: \((a^{2}b^{4})^{\frac{3}{2}}=a^{2\times\frac{3}{2}}b^{4\times\frac{3}{2}}=a^{3}b^{6}\).
Denominator: \((a^{-2}b^{-\frac{1}{3}})^{\frac{1}{3}}=a^{-2\times\frac{1}{3}}b^{-\frac{1}{3}\times\frac{1}{3}}=a^{-\frac{2}{3}}b^{-\frac{1}{9}}\).

Step 2: Divide using quotient rule for exponents

\(\frac{a^{3}b^{6}}{a^{-\frac{2}{3}}b^{-\frac{1}{9}}}=a^{3-(-\frac{2}{3})}b^{6-(-\frac{1}{9})}=a^{\frac{11}{3}}b^{\frac{55}{9}}\).

Problem 5:

Step 1: Simplify each term using power of a product rule

First term: \((\frac{1}{2}m^{-2}n^{\frac{1}{2}}p^{4})^{-2}=(\frac{1}{2})^{-2}(m^{-2})^{-2}(n^{\frac{1}{2}})^{-2}(p^{4})^{-2}=4m^{4}n^{-1}p^{-8}\).
Second term: \((m^{2}n^{\frac{3}{2}}p^{0})^{2}=m^{4}n^{3}p^{0}=m^{4}n^{3}\) (since \(p^{0}=1\)).

Step 2: Multiply the two results

Multiply \(4m^{4}n^{-1}p^{-8}\) and \(m^{4}n^{3}\):
\(4 \cdot m^{4 + 4} \cdot n^{-1+3} \cdot p^{-8}=4m^{8}n^{2}p^{-8}=\frac{4m^{8}n^{2}}{p^{8}}\).

Answer:

s:

1. \(\boldsymbol{\frac{1}{9}x^{2}y^{14}}\)
2. \(\boldsymbol{a^{\frac{11}{3}}b^{\frac{55}{9}}}\)
3. \(\boldsymbol{\frac{4m^{8}n^{2}}{p^{8}}}\)