Question

$\large\red{\sf{Question}}$

$\large\underline{\sf{If \: {2}^{x} = {3}^{y} = {6}^{z} \: \: then \: z = }}$
$\large\boxed{\sf{ \frac{x + y}{xy} }}$
$\large\boxed{\sf{ \frac{xy}{x + y} }}$
$\large\boxed{\sf{2xy}}$
$\large\boxed{\sf{ \frac{x}{y} }}$
​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

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

Let assume that

$\rm \: {2}^{x} = {3}^{y} = {6}^{z} = a$

So,

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

Also,

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

Also,

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

can be further rewritten as

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

On substituting the values of 2 and 3, from equation (1) and (2), we get

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

We know,

$\boxed{ \rm{ \: {x}^{m} \times {x}^{n} = {x}^{m + n} \: \: }}$

So, using this identity, we get

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

We know,

$\boxed{ \rm{ \: {x}^{m} \: = \: {x}^{n} \rm\implies \: m= n \: \: }}$

So, using this identity, we get

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

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

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

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

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