Question

please check attachment & explain​

Answer 1

Hello Mate,

$\huge{\blue{\star}} {\huge{\mathfrak{\red{\underline{Solution}}}}} \huge{\blue{\star}}$

Take any constant, say 'k'.

Let,

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

Now,

${2}^{x} = k \: \: \\ 2 = {k}^{ \frac{1}{x} } \: \: \: \: \: ....(1) \\ \\ {3}^{y} = k \\ 3 = {k}^{ \frac{1}{y} } \: \: \: \: \: ....(2)\\ \\ {6}^{ - z} = k \\ 6 = {k }^{ \frac{ - 1}{z} } \: \: \: \: ....(3)$

We know 2 x 3 = 6

By taking a look at the above three equations we can say,

${k}^{ \frac{1}{x} } \times {k}^{ \frac{1}{y} } = {k}^{ \frac{ - 1}{z} } \\ \\ as \: \: bases \: \: are \: \: equal \: \: equate \: \: the \: \: exponents \\ \\ \frac{1}{x} + \frac{1}{y} = - \frac{1}{z} \\ \\ \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0$

Hence Proved,

Hope it helps:)

Answer 2

I hope this will help you

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