Question

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

Answer 1

Sᴏʟᴜᴛɪᴏɴ :-

Let us Assume that, 2^x=3^y=6^(-z) = k . where k is any constant Integer. (k ≠ 0) .

Than,

→ 2^x = k

→ 2 = k^(1/x)

Similarly,

→ 3^y = k

→ 3 = k^(1/y)

And,

→ 6^(-z) = k

→ 6 = k^(-1/z)

Now, since , 2 * 3 = 6 .

Putting values we get,

→ k^(1/x) * k^(1/y) = k^(-1/z)

using a^m * a^n = a^(m + n) in LHS,

→ k^(1/x + 1/y) = k^(-1/z)

Now, using , a^m = a^n => m = n

→ (1/x + 1/y) = (-1/z)

→ (1/x + 1/y + 1/z) = 0 (Ans.)

Answer 2

${\huge{\bf{\red{\underline{Solution:}}}}}$

${\bf{\blue{\underline{Given:}}}}$

$\star{\sf{ \green{ \: \: {2}^{x} = {3}^{y} = {6}^{ - z} }}}$

${\bf{\blue{\underline{To\:Find:}}}}$

$\star{\sf{ \green{ \: \: \frac{1}{x} + \frac{1}{y} + \frac{1}{z} }}}$

${\bf{\blue{\underline{Now:}}}}$

Let ,

$\implies{\sf{ \: \: {2}^{x} = {3}^{y} = {6}^{ - z} = k(any \: constant)}}$

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$: \implies{\sf{ {2}^{x} = k }}$

$: \implies{\sf{ {2}= {k}^{ \frac{1}{x} } }}$

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$: \implies{\sf{ {3}^{y} = k }}$

$: \implies{\sf{ {3}= {k}^{ \frac{1}{y} } }}$

__________________________________

$: \implies{\sf{ {6}^{z} = k }}$

$: \implies{\sf{ {6}= {k}^{ \frac{-1}{z} } }}$

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Since,

$: \implies{\sf{ 2 \times 3 = 6}}$

Put all the the values in this,

$: \implies{\sf{ {k}^{ \frac{1}{x} } + {k}^{ \frac{ 1}{y} } = {k}^{ \frac{-1}{ z} } }}$

We know that ,

$\bigstar \: \: \underline{ \purple{ \boxed{ \sf{ {a}^{ \frac{p}{q} } + {a}^{ \frac{r}{s} } = {a}^{ \frac{p}{q} + \frac{r}{s} } }}}}$

$: \implies{\sf{ {(k})^{ \frac{1}{x} + \frac{1}{y} } = {k}^{ \frac{-1}{ z} } }}$

$\star \: {\sf{ Base \: same}}$

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

$\: {\implies{ \frac{1}{x} + \frac{1}{y} + \frac{ 1}{ z} }}$

Hence proved!