[tex]\large\underline{\sf{Solution-}}[/tex]
Given that,
[tex]\sf \: {3}^{x} = {4}^{y} = {12}^{z} \\ \\ [/tex]
Let assume that
[tex]\sf \: {3}^{x} = {4}^{y} = {12}^{z} = k \\ \\ [/tex]
Now,
[tex]\sf \: {3}^{x} = k \: \: \sf \: \implies \: 3 = {\bigg(k\bigg) }^{\dfrac{1}{x} } \\ \\ [/tex]
and
[tex]\sf \: {4}^{y} = k \: \: \sf \: \implies \: 4 = {\bigg(k\bigg) }^{\dfrac{1}{y} } \\ \\ [/tex]
Also,
[tex]\sf \: {12}^{z} = k \\ \\ [/tex]
[tex]\sf \: 12 = {\bigg(k\bigg) }^{\dfrac{1}{z} } \\ \\ [/tex]
[tex]\sf \: 4 \times 3 = {\bigg(k\bigg) }^{\dfrac{1}{z} } \\ \\ [/tex]
[tex]\sf \: {\bigg(k\bigg) }^{\dfrac{1}{x} } \times {\bigg(k\bigg) }^{\dfrac{1}{y} } = {\bigg(k\bigg) }^{\dfrac{1}{z} } \\ \\ [/tex]
[tex]\sf \: {\bigg(k\bigg) }^{\dfrac{1}{x} + \dfrac{1}{y} } = {\bigg(k\bigg) }^{\dfrac{1}{z} } \\ \\ [/tex]
[tex]\sf \: \dfrac{1}{x} + \dfrac{1}{y} = \dfrac{1}{z} \\ \\ [/tex]
[tex]\sf \: \dfrac{y + x}{xy} = \dfrac{1}{z} \\ \\ [/tex]
[tex]\bf\implies \:z(x + y) = xy \\ \\ [/tex]
Hence, Proved
[tex]\rule{190pt}{2pt}[/tex]
Additional Information
[tex]\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}[/tex]
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Verified answer
[tex]\large\underline{\sf{Solution-}}[/tex]
Given that,
[tex]\sf \: {3}^{x} = {4}^{y} = {12}^{z} \\ \\ [/tex]
Let assume that
[tex]\sf \: {3}^{x} = {4}^{y} = {12}^{z} = k \\ \\ [/tex]
Now,
[tex]\sf \: {3}^{x} = k \: \: \sf \: \implies \: 3 = {\bigg(k\bigg) }^{\dfrac{1}{x} } \\ \\ [/tex]
and
[tex]\sf \: {4}^{y} = k \: \: \sf \: \implies \: 4 = {\bigg(k\bigg) }^{\dfrac{1}{y} } \\ \\ [/tex]
Also,
[tex]\sf \: {12}^{z} = k \\ \\ [/tex]
[tex]\sf \: 12 = {\bigg(k\bigg) }^{\dfrac{1}{z} } \\ \\ [/tex]
[tex]\sf \: 4 \times 3 = {\bigg(k\bigg) }^{\dfrac{1}{z} } \\ \\ [/tex]
[tex]\sf \: {\bigg(k\bigg) }^{\dfrac{1}{x} } \times {\bigg(k\bigg) }^{\dfrac{1}{y} } = {\bigg(k\bigg) }^{\dfrac{1}{z} } \\ \\ [/tex]
[tex]\sf \: {\bigg(k\bigg) }^{\dfrac{1}{x} + \dfrac{1}{y} } = {\bigg(k\bigg) }^{\dfrac{1}{z} } \\ \\ [/tex]
[tex]\sf \: \dfrac{1}{x} + \dfrac{1}{y} = \dfrac{1}{z} \\ \\ [/tex]
[tex]\sf \: \dfrac{y + x}{xy} = \dfrac{1}{z} \\ \\ [/tex]
[tex]\bf\implies \:z(x + y) = xy \\ \\ [/tex]
Hence, Proved
[tex]\rule{190pt}{2pt}[/tex]
Additional Information
[tex]\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}[/tex]