Question

If 2x=3y=6-z, prove that 1/x+1/y+1/z =0​

Answer 1

Correct Statement is

$\sf \: If \: {2}^{x} = {3}^{y} = {6}^{ - z}, \: prove \: that \: \dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} = 0$

Answer

Identities Used :-

$\boxed{ \red{ \sf \: If \: {x}^{y} = a, \: then \: x = {\bigg(a \bigg)}^{\dfrac{1}{y} }}}$

$\boxed{ \red{ \sf \: {a}^{x} = {a}^{y} \: \: \: then \: \: x \: = \: y}}$

Step by step solution

$\rm :\longmapsto\:Let \: \: {2}^{x} = {3}^{y} = {6}^{ - z} = k$

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

↝ Aɢᴀɪɴ,

$\rm :\implies\: {3}^{y} = k \: \implies \: 3 = {\bigg(k \bigg)}^{ \dfrac{1}{y}} - - - (2)$

↝ Aɢᴀɪɴ,

$\rm :\implies\: {6}^{ - z} = k \: \implies \: 6 = {\bigg(k \bigg)}^{ \dfrac{ - 1}{z}} - - - (3)$

↝ Now,

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

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

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

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

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

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

${\boxed{\boxed{\bf{Hence, Proved}}}}$

Additional Information :-

$\begin{gathered}(1)\:{\underline{\boxed{\bf{\blue{a^m\times{a^n}\:=\:a^{m\:+\:n}\:}}}}} \\ \end{gathered}$

$\begin{gathered}(2)\:{\underline{\boxed{\bf{\purple{\dfrac{a^m}{a^n}\:=\:a^{m\:-\:n}\:}}}}} \\ \end{gathered}$

$\begin{gathered}(3)\:{\underline{\boxed{\bf{\orange{\dfrac{1}{x^n}\:=\:x^{-n}\:}}}}} \\ \end{gathered}$

$\begin{gathered}(4)\:{\underline{\boxed{\bf{\color{peru}{(a^m)^n\:=\:a^{m\times{n}}\:}}}}} \\ \end{gathered}$

$\begin{gathered}(5)\:{\underline{\boxed{\bf{\red{ {x}^{0} = 1}}}}} \\ \end{gathered}$