Question

Which one among the following is correct?

• A. ${\left(\frac{a}{b}\right)}^m=a^m{b}^m$
• B. ${\left(\frac{a}{b}\right)}^{-m}$ = $b^m$$a^m$
• C. ${\left(\frac{a}{b}\right)}^m$ = $\frac{\left(b^m\right)}{\left(a^m\right)}$
• D. ${\left(\frac{b}{a}\right)}^{-m}$ = $b^{-m}{a}^m$

Answer 1

The correct option is D

${\left(\frac{b}{a}\right)}^{-m}$ = $b^{-m}{a}^m$

Consider $(\frac{a}{b}{)}^m$,

$(\frac{a}{b}{)}^m=\frac{a^m}{b^m}\ne \frac{b^m}{a^m}$

$(\frac{a}{b}{)}^m=\frac{a^m}{b^m}[\because (\frac{x}{y}{)}^r=\frac{x^r}{y^r}]=a^m\times b^{-m}[\because \frac{1}{x^n}=x^{-n}]\ne a^m{b}^m$

Consider $(\frac{b}{a}{)}^{-m}$,

$(\frac{b}{a}{)}^{-m}=\frac{b^{-m}}{a^{-m}}[\because (\frac{x}{y}{)}^r=\frac{x^r}{y^r}]=b^{-m}{a}^m[\because \frac{1}{x^{-n}}=x^n]$

Consider $(\frac{a}{b}{)}^{-m}$,

$(\frac{a}{b}{)}^{-m}=\frac{a^{-m}}{b^{-m}}[\because (\frac{x}{y}{)}^r=\frac{x^r}{y^r}]=a^{-m}\times b^m[\because \frac{1}{x^{-n}}=x^n]\ne a^m{b}^m$

$\therefore (\frac{b}{a}{)}^{-m}=b^{-m}{a}^m$