Question

If $a^7=3$, find the value of $\frac{{\left(a^{-2}\right)}^{-3}\times {\left(a^3\right)}^4\times {\left(a^{-17}\right)}^{-1}}{a^7}$

Answer 1

Consider, $\frac{(a^{-2}{)}^{-3}\times (a^3{)}^4\times (a^{-17}{)}^{-1}}{a^7}$

$=\frac{(a{)}^{\left[\right(-2)\times (-3\left)\right]}\times (a{)}^{(3\times 4)}\times (a{)}^{(-17)\times (-1)}}{a^7}$ ...... [Using power rule of exponents]

$=\frac{a^6\times a^{12}\times a^{17}}{a^7}$

$=\frac{a^{(6+12+17)}}{a^7}$ ........ [Using product rule of exponents]

$=\frac{a^{35}}{a^7}$

$=a^{35-7}$ ............ [Using quotient rule of exponents]

$=a^{28}$

$=(a^7{)}^4$ ........ [Using power rule of exponents]

$=(3{)}^4$ ......... [Given]

$=81$

Hence, the correct answer is $81$.