Question

Simplify ${49}^{-5}\times {49}^3$.

• A. $7^4$
• B. ${49}^{-5}$
• C. ${49}^{-2}$
• D. $\frac{1}{7^4}$

Answer 1

Answer: (C), (D)

The correct options are
C ${49}^{-2}$
D $\frac{1}{7^4}$

${49}^{-5}\times {49}^3$
$={49}^{(-5+3)}$ $(\because a^m\times a^n=a^{m+n})$
$={49}^{-2}$

Also, by using $(a^m{)}^n=a^{mn}$, we get
${49}^{-2}$ = $(7^2{)}^{-2}=(7{)}^{2\times -2}=(7{)}^{-4}$

But $(7{)}^{-4}=\frac{1}{7^4}$ $[\because a^{-m}=\frac{1}{a^m}]$

$\therefore {49}^{-5}\times {49}^3={49}^{-2}=\frac{1}{7^4}$