Question

\[ \frac{(a^{-1}b^2 / a^2 b^{-4})^7}{(a^3 b^{-5} / a^{-2} b^3)} = a^x b^y \]

Solution

\[ = \frac{(a^{-1-2} \cdot b^{2-(-4)})^7}{a^{3-(-2)} \cdot b^{-5-3}} \] \[ = \frac{(a^{-3} \cdot b^{6})^7}{a^{5} \cdot b^{-8}} \] \[ = \frac{a^{-21} \cdot b^{42}}{a^{5} \cdot b^{-8}} \] \[ = a^{-21-5} \cdot b^{42-(-8)} \] \[ = a^{-26} \cdot b^{50} \] \[ x = -26,\quad y = 50 \]

Answer

\[ \boxed{x = -26,\quad y = 50} \]

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