Question

if 2^x+3^y+6^-z ,find the value of 1/x+1/y+1/z​

Answer 1

Answer:

Given,

${2}^{x} + {3}^{y} + {6}^{ - z}$

we assumed that,

$\: \: \: \: \sf \: {2}^{x} = {3}^{y} = {6}^{ - z} = k \\ \sf \therefore \: {2}^{x} = k \: \\ = > \sf \: 2 = {k}^{ \frac{1}{x} } \: \: \: \: - - - - (1) \\ \\ \sf \therefore \: \: {3}^{y} = k \\ = > \sf \: 3 = {k}^{ \frac{1}{y} } \: \: \: \: - - - - (2) \\ \\ \sf \therefore \: {6}^{ - z} = k \\ \sf \: = > \frac{1}{6} = {k}^{ \frac{1}{z} } \: \: \: \: \: - - - - (3) \\ \\ (1) \times (2) \times (3) \: \: \: \sf \sf \: \implies \: \: \\ \sf \: {k}^{ \frac{1}{x} } \times {k}^{ \frac{1}{y} } \times {k}^{ \frac{1}{z} } = 2 \times 3 \times \frac{1}{6} = 1 \\ \sf \: \implies \: {k}^{( \frac{1}{x} + \frac{1}{y} + \frac{1}{z} )} = 1 = {k}^{0} \\ \sf \therefore \: \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0$

Note:

$> \sf {a}^{m} = {a}^{n} \: \: \: \: \: \: \sf \therefore \: \: m \: = n \: \: \: \: \: \: \\: \\ > {x}^{0} = 1 \: \: \: \: \{ \: x \neq0 \: \}$