Question

If 2^{x} = 3^{y} = 6^{ - z} then x^{-1}+y^{-1}+z^{-1} is equal to​

Answer 1

Answer:

$\boxed{\bf \: {x}^{ - 1} + {y}^{ - 1} + {z}^{ - 1} = 0 \: }$

Step-by-step explanation:

Given that,

$\sf \: {2}^{x} = {3}^{y} = {6}^{- z}$

Let assume that

$\sf \: {2}^{x} = {3}^{y} = {6}^{- z} = k$

Now,

$\sf \: {2}^{x} = k \: \: \sf \:  \implies \: 2 = {\bigg(k\bigg) }^{\dfrac{1}{x} }$

Also,

$\sf \: {3}^{y} = k \: \: \sf \:  \implies \: 3 = {\bigg(k\bigg) }^{\dfrac{1}{y} }$

Also,

$\sf \: {6}^{- z} = k$

$\sf \: 6 = {\bigg(k\bigg) }^{- \dfrac{1}{z} }$

$\sf \: 2 \times 3 = {\bigg(k\bigg) }^{- \dfrac{1}{z} }$

$\sf \: {\bigg(k\bigg) }^{\dfrac{1}{x} } \times {\bigg(k\bigg) }^{\dfrac{1}{y} } = {\bigg(k\bigg) }^{- \dfrac{1}{z} }$

$\sf \: {\bigg(k\bigg) }^{\dfrac{1}{x} + \dfrac{1}{y} } = {\bigg(k\bigg) }^{- \dfrac{1}{z} }$

$\sf \: \dfrac{1}{x} + \dfrac{1}{y} = - \dfrac{1}{z}$

$\sf \: \dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} = 0$

$\implies\sf \: {x}^{ - 1} + {y}^{ - 1} + {z}^{ - 1} = 0$

Hence,

$\implies\sf \: \boxed{\bf \: {x}^{ - 1} + {y}^{ - 1} + {z}^{ - 1} = 0 \: }$

$\rule{190pt}{2pt}$

Additional Information

$\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{{More \: Formulae}}}} \\ \\ \bigstar \: \bf{ {x}^{0} = 1}\\ \\ \bigstar \: \bf{ {x}^{m} \times {x}^{n} = {x}^{m + n} }\\ \\ \bigstar \: \bf{ {( {x}^{m})}^{n} = {x}^{mn} }\\ \\\bigstar \: \bf{ {x}^{m} \div {x}^{n} = {x}^{m - n} }\\ \\ \bigstar \: \bf{ {x}^{ - n} = \dfrac{1}{ {x}^{n} } }\\ \\\bigstar \: \bf{ {\bigg(\dfrac{a}{b} \bigg) }^{ - n} = {\bigg(\dfrac{b}{a} \bigg) }^{n} }\\ \\\bigstar \: \bf{ {x}^{m} = {x}^{n}\rm\implies \:m = n }\\ \\ \end{array} }}\end{gathered}\end{gathered}\end{gathered}$