Q. 5.41 - Intermediate Algebra

Question

Simplify each expression: 

$ⓐ {\left(c^4{d}^2\right)}^5 {\left(3c{d}^5\right)}^4ⓑ \frac{\left(a-2\right)3 \left(a2\right)4 }{\left(a4\right)5} ⓒ {\left(\frac{3x{y}^2}{x^2{y}^{-3}}\right)}^2 {\left(\frac{9x{y}^{-3}}{x^3{y}^2}\right)}^{-1}$

Step-by-Step Solution

Answer

The simplified expression of ${\left(c^4{d}^2\right)}^5 {\left(3c{d}^5\right)}^4=81c^{24}{d}^{30}$

The simplified expression of $\frac{{\left(a^{-2}\right)}^3 {\left(a^2\right)}^4 }{{\left(a^4\right)}^5}=\frac{1}{a^{18}}$

The simplified expression of ${\left(\frac{3x{y}^2}{x^2{y}^{-3}}\right)}^2 {\left(\frac{9x{y}^{-3}}{x^3{y}^2}\right)}^{-1}=9y^{15}$

Step 1. Given Information

In the given question we have to simplify each expression: 

$ⓐ {\left(c^4{d}^2\right)}^5 {\left(3c{d}^5\right)}^4ⓑ \frac{{\left(a^{-2}\right)}^3 {\left(a^2\right)}^4 }{{\left(a^4\right)}^5} ⓒ {\left(\frac{3x{y}^2}{x^2{y}^{-3}}\right)}^2 {\left(\frac{9x{y}^{-3}}{x^3{y}^2}\right)}^{-1}$

Part (a) Step 1. The given expression is ( c 4 d 2 ) 5   ( 3 c d 5 ) 4

Use the Product to a Power Property, ${\left(ab\right)}^m=a^m{b}^m$

${\left(c^4{d}^2\right)}^5 {\left(3c{d}^5\right)}^4={{\left(c^4\right)}^5\left(d^2\right)}^5 {{\left(3\right)}^4{\left(c\right)}^4\left(d^5\right)}^4$

${\left(c^4{d}^2\right)}^5 {\left(3c{d}^5\right)}^4=\left(c^{4·5}\right)\left(d^{2·5}\right) {\left(3\right)}^4{\left(c\right)}^4\left(d^{5·4}\right)$

Simplify.

${\left(c^4{d}^2\right)}^5 {\left(3c{d}^5\right)}^4=\left(c^{20}\right)\left(d^{10}\right) 81c^4\left(d^{20}\right)$

Use the Commutative Property.

${\left(c^4{d}^2\right)}^5 {\left(3c{d}^5\right)}^4=81c^{4+20}{d}^{20+10}$

Multiply the constants and add the exponents.

${\left(c^4{d}^2\right)}^5 {\left(3c{d}^5\right)}^4=81c^{24}{d}^{30}$

Part (b) Step 1. The given expression is ( a − 2 ) 3   ( a 2 ) 4   ( a 4 ) 5

Use the Power Property, ${\left(a^m\right)}^n=a^{m·n}$

$\frac{{\left(a^{-2}\right)}^3 {\left(a^2\right)}^4 }{{\left(a^4\right)}^5}=\frac{\left(a^{-2·3}\right)\left(a^{2·4}\right) }{\left(a^{4·5}\right)}$

$\frac{{\left(a^{-2}\right)}^3 {\left(a^2\right)}^4 }{{\left(a^4\right)}^5}=\frac{\left(a^{-6}\right)\left(a^8\right) }{\left(a^{20}\right)}$

Add the exponents in the numerator.

$\frac{{\left(a^{-2}\right)}^3 {\left(a^2\right)}^4 }{{\left(a^4\right)}^5}=\frac{a^{-6+8} }{a^{20}}$

$\frac{{\left(a^{-2}\right)}^3 {\left(a^2\right)}^4 }{{\left(a^4\right)}^5}=\frac{a^2 }{a^{20}}$

Use the Quotient Property $\frac{a^m}{a^n}=\frac{1}{a^{n-m}}$

$\frac{{\left(a^{-2}\right)}^3 {\left(a^2\right)}^4 }{{\left(a^4\right)}^5}=\frac{1}{a^{20-2}}$

$\frac{{\left(a^{-2}\right)}^3 {\left(a^2\right)}^4 }{{\left(a^4\right)}^5}=\frac{1}{a^{18}}$

Part (c) Step 1. The given expression is 3 x y 2 x 2 y − 3 2   9 x y − 3 x 3 y 2 − 1

Simplify inside the parentheses first.

${\left(\frac{3x{y}^2}{x^2{y}^{-3}}\right)}^2 {\left(\frac{9x{y}^{-3}}{x^3{y}^2}\right)}^{-1}={\left(\frac{3y^5}{x}\right)}^2·{\left(\frac{9}{x^2{y}^5}\right)}^{-1}$

Use the Quotient to a Power Property, ${\left(\frac{a}{b}\right)}^m=\frac{a^m}{b^m}$

${\left(\frac{3x{y}^2}{x^2{y}^{-3}}\right)}^2 {\left(\frac{9x{y}^{-3}}{x^3{y}^2}\right)}^{-1}=\frac{{\left(3\right)}^2{\left(y^5\right)}^2}{{\left(x\right)}^2}·\frac{{\left(9\right)}^{-1}}{{\left(x^2\right)}^{-1}{\left(y^5\right)}^{-1}}$

${\left(\frac{3x{y}^2}{x^2{y}^{-3}}\right)}^2 {\left(\frac{9x{y}^{-3}}{x^3{y}^2}\right)}^{-1}=\frac{9{\left(y\right)}^{10}}{{\left(x\right)}^2}·\frac{{\left(9\right)}^{-1}}{{\left(x\right)}^{-2}{\left(y\right)}^{-5}}$

Part (d) Step 2. Use the Product to a Power Property, ( a b ) m = a m b m

Simplify.

${\left(\frac{3x{y}^2}{x^2{y}^{-3}}\right)}^2 {\left(\frac{9x{y}^{-3}}{x^3{y}^2}\right)}^{-1}=\frac{9{\left(y\right)}^{10}}{{\left(x\right)}^2}·\frac{x^2{y}^5}{9}$

Simplify.

${\left(\frac{3x{y}^2}{x^2{y}^{-3}}\right)}^2 {\left(\frac{9x{y}^{-3}}{x^3{y}^2}\right)}^{-1}=9{\left(y\right)}^{10}·y^5$

${\left(\frac{3x{y}^2}{x^2{y}^{-3}}\right)}^2 {\left(\frac{9x{y}^{-3}}{x^3{y}^2}\right)}^{-1}=9y^{10+5}$

${\left(\frac{3x{y}^2}{x^2{y}^{-3}}\right)}^2 {\left(\frac{9x{y}^{-3}}{x^3{y}^2}\right)}^{-1}=9y^{15}$