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Laws of Exponents for Real Numbers

If a = xyp-1,...

Question

If a = $x y^{p - 1}, b = x y^{q - 1} a n d c = x y^{r - 1}$ , prove that $a^{q - r} b^{r - p} c^{p - q}$ = 1,

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Solution

$a = x y^{p - 1}, b = x y^{q - 1} a n d c = x y^{r - 1} a^{q - r} b^{r - p} c^{p - q} = 1 L H S = a^{q - r} b^{r - p} c^{p - q} = (x y^{p - 1})^{q - r} . (x y^{q - 1})^{r - p} . (x y^{r - 1})^{p - q} = x^{q - r} . y^{(p - 1) (q - r)} . x^{r - p} . y^{(q - 1) (r - p)} . x^{p - q} . y^{(r - 1) (p - q)} = x^{q - r + r - p + p - q} . y^{(p - 1) (q - r)} . y^{(q - 1) (r - p)} . y^{(r - 1) (p - q)} = x^{0} . y^{p q - p r - q + r + q r - p q - r + p + r p - r q - p + q} = x^{0} . y^{0} = 1 \times 1 = 1 = R H S$ $⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} ∵ a^{0} = 1 (a^{m})^{n} = a^{m n} a^{m} \times a^{n} = a^{m + n} \end{matrix} ⎫ ⎪ ⎬ ⎪ ⎭$

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a = x y^{p - 1}, b = x y^{q - 1}

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