Question

If 2^x=3^y=6^z .then find 1/x+1/y+1/z
$if \: {2}^{x} = {3}^{y} = {6}^{z} .then \: find \: \frac{1}{x} + \frac{1}{y} + \frac{1}{z}$
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Answer 1

Step-by-step explanation:

In order to find the required result, given question requires an improvement.

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

should be expressed as:

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

If it is so, then let us find:

$Given: \: {2}^{x} = {3}^{y} = {6}^{ - z} \\ Let \: {2}^{x} = {3}^{y} = {6}^{ - z} = k \\ \implies \: {2}^{x} =k\implies 2 = k^{ \frac{1}{x} } \\ \implies \: {3}^{y} =k\implies 3= k^{ \frac{1}{y} } \\ \implies \: {6}^{ - z} =k\implies 6= k^{ - \frac{1}{z} } \\ now \\ \because \: \: 2 \times 3 = 6 \\ \therefore \: k^{ \frac{1}{x} } \times k^{ \frac{1}{y} } = k^{ - \frac{1}{z} } \\ \implies \: k^{ \frac{1}{x} + \frac{1}{y}}= k^{ - \frac{1}{z} } \\ \implies \: \frac{1}{x} + \frac{1}{y} = - \frac{1}{z} \\ \implies \: \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0$

Answer 2

Step-by-step explanation:

In order to find the required result, given question requires an improvement.

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

should be expressed as:

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

If it is so, then let us find:

$Given: \: {2}^{x} = {3}^{y} = {6}^{ - z} \\ Let \: {2}^{x} = {3}^{y} = {6}^{ - z} = k \\ \implies \: {2}^{x} =k\implies 2 = k^{ \frac{1}{x} } \\ \implies \: {3}^{y} =k\implies 3= k^{ \frac{1}{y} } \\ \implies \: {6}^{ - z} =k\implies 6= k^{ - \frac{1}{z} } \\ now \\ \because \: \: 2 \times 3 = 6 \\ \therefore \: k^{ \frac{1}{x} } \times k^{ \frac{1}{y} } = k^{ - \frac{1}{z} } \\ \implies \: k^{ \frac{1}{x} + \frac{1}{y}}= k^{ - \frac{1}{z} } \\ \implies \: \frac{1}{x} + \frac{1}{y} = - \frac{1}{z} \\ \implies \: \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0$

Answer 3

Step-by-step explanation:

In order to find the required result, given question requires an improvement.

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

should be expressed as:

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

If it is so, then let us find:

$Given: \: {2}^{x} = {3}^{y} = {6}^{ - z} \\ Let \: {2}^{x} = {3}^{y} = {6}^{ - z} = k \\ \implies \: {2}^{x} =k\implies 2 = k^{ \frac{1}{x} } \\ \implies \: {3}^{y} =k\implies 3= k^{ \frac{1}{y} } \\ \implies \: {6}^{ - z} =k\implies 6= k^{ - \frac{1}{z} } \\ now \\ \because \: \: 2 \times 3 = 6 \\ \therefore \: k^{ \frac{1}{x} } \times k^{ \frac{1}{y} } = k^{ - \frac{1}{z} } \\ \implies \: k^{ \frac{1}{x} + \frac{1}{y}}= k^{ - \frac{1}{z} } \\ \implies \: \frac{1}{x} + \frac{1}{y} = - \frac{1}{z} \\ \implies \: \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0$