Chapter 5.1, Problem 30WE

To determine

To find: Simplify the given expression.

Answer to Problem 30WE

  $\frac{a^{n-1}\cdot b^{2n}}{a^{n+1}\cdot {\left(b^2\right)}^{n-1}}=\frac{b}{a^2}$

Explanation of Solution

Given information:

  $\frac{a^{n-1}\cdot b^{2n}}{a^{n+1}\cdot {\left(b^2\right)}^{n-1}}$

Concept:

  $\frac{a}{b}\cdot \frac{c}{d}=\frac{a\cdot c}{b\cdot d}$

Using law of exponents

  $\begin{array}{l}\frac{a^m}{a^n}=a^{m-n}\\ a^m\cdot a^n=a^{m+n}\end{array}$

  ${\left(a^n\right)}^m=a^{mn}$

Where $a$ be any value.

  $\frac{a^{n-1}\cdot b^{2n}}{a^{n+1}\cdot {\left(b^2\right)}^{n-1}}$

  $\Rightarrow \frac{a^{n-1}\cdot b^{2n}}{a^{n+1}\cdot {\left(b^2\right)}^{n-1}}=\frac{a^{n-1}\cdot b^{2n}}{a^{n+1}\cdot b^{2\left(n-1\right)}}$

  $\Rightarrow \frac{a^{n-1}\cdot b^{2n}}{a^{n+1}\cdot {\left(b^2\right)}^{n-1}}=\frac{a^{n-1}\cdot b^{2n}}{a^{n+1}\cdot b^{2n-2}}$

  $\Rightarrow \frac{a^{n-1}\cdot b^{2n}}{a^{n+1}\cdot {\left(b^2\right)}^{n-1}}=\frac{a^{n-1}}{a^{n+1}}\cdot \frac{b^{2n}}{b^{2n-1}}$

  $\Rightarrow \frac{a^{n-1}\cdot b^{2n}}{a^{n+1}\cdot {\left(b^2\right)}^{n-1}}=a^{n-1-\left(n+1\right)}\cdot b^{2n-\left(2n-1\right)}$

  $\Rightarrow \frac{a^{n-1}\cdot b^{2n}}{a^{n+1}\cdot {\left(b^2\right)}^{n-1}}=a^{n-1-n-1}\cdot b^{2n-2n+1}$

  $\Rightarrow \frac{a^{n-1}\cdot b^{2n}}{a^{n+1}\cdot {\left(b^2\right)}^{n-1}}=a^{-2}\cdot b^1$

  $\Rightarrow \frac{a^{n-1}\cdot b^{2n}}{a^{n+1}\cdot {\left(b^2\right)}^{n-1}}=\frac{b}{a^2}$

Therefore,

  $\frac{a^{n-1}\cdot b^{2n}}{a^{n+1}\cdot {\left(b^2\right)}^{n-1}}=\frac{b}{a^2}$