Question

Do if you know.

Maths : Indices​

Answer 1

Answer:

Step-by-step explanation:

We have,

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

$\sf{\implies\,2=k^{\dfrac{1}{x}}\,\,\,\,3=k^{\dfrac{1}{y}}\,\,\,\,6=k^{-\dfrac{1}{z}}}$

Now,

$\sf{\dfrac{2\times3}{6}=\dfrac{k^{\dfrac{1}{x}}\cdot\,k^{\dfrac{1}{y}}}{k^{-\dfrac{1}{z}}}}$

$\sf{\implies\,\dfrac{6}{6}=k^{\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}}}$

$\sf{\implies\,k^{\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}}=1}$

$\sf{\implies\,k^{\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}}=k^{0}}$

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

Answer 2

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

Given that,

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

Let assume that

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

So,

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

Again,

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

Also,

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

Now,

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

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

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

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

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

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

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Identities Used :-

$\boxed{ \tt{ \: {x}^{y} = z \: \rm \implies\:x = {\bigg(z\bigg) }^{\dfrac{1}{y} }}}$

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

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More to know :-

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

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

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