Question

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Answer 1

$\boxed{\mathsf{ Solution : \: }}$

$\mathtt{ Let , \: } \\ \\ \mathtt{ \: \implies{{2}^{x} \: = \: {3}^{y} \: = \: {6}^{ - z} \: = \: k}} \\ \\ \mathtt{ \: \implies{2 \: = {k}^{ \frac{1}{x} } }}\\ \\ \mathtt{\implies{ \: 3 \: = \: {k}^{ \frac{1}{y} } }}$

$\mathsf{ Now, \: }$

$\mathtt{ \: \implies{\: {6}^{ - z} \: = \: k }} \\ \\ \mathtt{ \: \implies{ \: ( \: 2 \: \times \: 3 \: )^{ - z} \: = \: k }}$

$\mathsf{ Plug \: the \: value \: of \: 2 \: and \: 3 , \: }$

$\mathtt{\implies{( \: \: {k}^{ \frac{1}{x} } \: \times \: {k}^{ \frac{1}{y} } \: )^{ - z} \: = \: k }}$

$\mathsf{ Using \: \: algebraic \: \: identity , \: }$

$\boxed{ \mathtt{ \: \implies{ \: {a}^{b} \: \times \: {a}^{c} \: = \: {a}^{( \: b \: + \: c \: )} }} }$

$\mathtt{ \: \implies{ \: ( \: {k}^{ \frac{1}{x} \: + \: \frac{1}{y} } \: ) {}^{ - z} \: = \: k \: }}$

$\mathsf{ Using \: \: algebraic \: \: identity, \: }$

$\boxed{ \mathtt{\implies{( {a}^{b} ) {}^{c} \: = \: {a}^{(bc)}} }}$

$\mathtt{\implies{ {k}^{( \: \frac{1}{x} \: + \: \frac{1}{y} )( - z)} \: = \: {k}^{1} } }$

$\mathsf{ On \: \: Comparison , \: }$

$\mathtt{ \implies{ \: ( \: \frac{1}{x} \: + \: \frac{1}{y} \: )( \: - z \: ) \: = \: 1 \: }}$

$\mathtt{ \: \implies{ \: ( \: \frac{1}{x} \: + \: \frac{1}{y} \: )\: = \: \frac{1}{ - z} }} \\ \\ \mathtt{\implies{ \: \frac{1}{x} \: + \: \frac{1}{y} \: = \: \frac{ - 1}{z} }} \\ \\ \mathtt{\implies{ \: \frac{1}{x} \: + \: \frac{1}{y} \: + \: \frac{1}{z} \: = \: 0 }}$

$\boxed{\mathsf{ Proved \: \ !! \: }}$