Question

$if \: {2}^{x} = {7}^{ - y} = {14}^{z } \\ prove \: that \: \frac{1}{x } = \frac{1}{y} + \frac{1}{z}$

Answer 1

Heya!!

Here's your answer!!
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Given,
${2}^{x} = {7}^{ - y} = {14}^{z} \\ let \: {2}^{x} = {7}^{y} = {14}^{z} \\ = > {2}^{x} = k \: or \: 2 = {k}^{ \frac{1}{x} } \\ {7}^{ - y} = k \: \: \: or \: 7 = {k}^{ \frac{ - 1}{z} } \\ {14}^{z} = k \: or \: 14 = {k}^{ \frac{1}{z} } \\ now \: \\ 14 = 2 \times 7 = > {k}^{ \frac{1}{z} } = {k}^{ \frac{1}{x} } . {k}^{ \frac{ - 1}{y} } \\ = > {k}^{ \frac{1}{z} } = {k}^{ \frac{1}{x } - \frac{1}{y} } \\ \\ = > \frac{1}{z} = \frac{1}{x} - \frac{1}{y} = > \frac{1}{x} = \frac{1}{y} + \frac{1}{z}$

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