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Power of a Power

Prove that : ...

Question

Prove that : ${(\frac{2^{a}}{2^{b}})}^{(a + b)} . {(\frac{2^{b}}{2^{c}})}^{(b + c)} . {(\frac{2^{c}}{2^{d}})}^{(c + d)} . {(\frac{2^{d}}{2^{a}})}^{(d + a)} = 1$

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Solution

${(\frac{2^{a}}{2^{b}})}^{(a + b)} \cdot {(\frac{2^{b}}{2^{c}})}^{(b + c)} \cdot {(\frac{2^{c}}{2^{d}})}^{(c + d)} \cdot {(\frac{2^{d}}{2^{a}})}^{(d + a)}$

$= {(2^{(a - b)})}^{(a + b)} \cdot {(2^{(b - c)})}^{(b + c)} \cdot {(2^{(c - d)})}^{(c + d)} \cdot {(2^{(d - a)})}^{(d + a)}$

$= 2^{(a^{2} - b^{2})} \cdot 2^{(b^{2} - c^{2})} \cdot 2^{(c^{2} - d^{2})} \cdot 2^{(d^{2} - a^{2})}$

$= 2^{(a^{2} - b^{2} + b^{2} - c^{2} + c^{2} - d^{2} + d^{2} - a^{2})}$

$= 2^{0}$

$= 1$

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Similar questions

{(\frac{2^{a}}{2^{b}})}^{a + b} \times {(\frac{2^{b}}{2^{c}})}^{b + c} \times {(\frac{2^{c}}{2^{a}})}^{c + a} = 1

Q. The rate of a reaction is expressed in different ways as follows:

\frac{1}{2} (\frac{d [C]}{d t}) = - \frac{1}{3} (\frac{d [D]}{d t}) = \frac{1}{4} (\frac{d [A]}{d t}) = - (\frac{d [B]}{d t})

The reaction is

∣ ∣ ∣ ∣ \begin{matrix} a - b - c & 2 a & 2 a 2 b & b - c - a & 2 b 2 c & 2 c & c - a - b \end{matrix} ∣ ∣ ∣ ∣ = {(a + b + c)}^{3}

∣ ∣ ∣ ∣ \begin{matrix} a - b - c & 2 a & 2 a 2 b & b - c - a & 2 b 2 c & 2 c & c - a - b \end{matrix} ∣ ∣ ∣ ∣ = {(a + b + c)}^{3}

\frac{b - c}{a} {cos}^{2} \frac{1}{2} A + \frac{c - a}{b} {cos}^{2} \frac{1}{2} B + \frac{a - b}{c} {cos}^{2} \frac{1}{2} C = 0.

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