Question
$(-7)^{-4} \times(-7)^2=(-7)^{-2}$

Answer

$LHS =(-7)^{-4} \times(-7)^2$
Using law of exponents, $a^m \times a^n=(a)^{m+n}$ [$\because$ a is non-zero integer]
$\therefore(-7)^{-4} \times(-7)^2=(-7)^{-4+2}$
$=(-7)^{-2}$
$LHS ≠ RHS$

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