Question

If 2^x=3^y=6^z ,then z is equal to​

Answer 1

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

Given that,

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

Let assume that

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

So,

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

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

and

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

Now, we have

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

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

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

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

$\red{ \bigg\{ \sf \: \because \: {a}^{m} \times {a}^{n} = {a}^{m + n} \bigg\}}$

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

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

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

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EXPLORE MORE

$\boxed{\tt{ {a}^{m} \div {a}^{n} = {a}^{m - n} \: }}$

$\boxed{\tt{ {( {x}^{m} )}^{n} = {x}^{mn} \: }}$

$\boxed{\tt{ \: {x}^{0} = 1 \: }}$

$\boxed{\tt{ \: {x}^{ - y} = \frac{1}{ {x}^{y} } \: }}$

Answer 2

$\color{purple}{ \colorbox{orange}{\colorbox{white} {hope \: it \: helps \: you}}}$