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I am a grade 8 student and i am aware that $ \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0$ is proved by setting all values $= k$.
See the question number 24 in my assignment.

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4 Answers 4

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Actually you have because of misprint mistake in your question.

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

$ 2 = k^{\frac{1}{x}}$

$ 3 = k^{\frac{1}{y}}$

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

As we know,

2.3 = 6

$k^{\frac{1}{x}} \cdot k^{\frac{1}{y}} = k^{\frac{1}{-z}}$

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

On comparing powers we have,

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

$\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0$

To prove $\frac{1}{x} + \frac{1}{y} - \frac{1}{z} = 0$

We should have expression $2^x = 3^y = 6^z$

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    Ok so you are saying solving this question is not possible... – 2017-01-22
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    No I am not saying that. I am saying some typo in this question. – 2017-01-22
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    Maybe misprinting in question paper. – 2017-01-22
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    I have found similar question see https://in.answers.yahoo.com/question/index?qid=20120131065220AAt22t6 – 2017-01-22
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    Not exactly the same document, but obviously it's a simple reformatting of the one from the picture, [here](http://summerfields.co.in/wp-content/uploads/2015/12/ARYABHATTA-QP-VIII-2015.pdf). Undoubtedly misprinting. – 2017-01-22
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    You have taken this question from this paper? – 2017-01-22
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    Yes question paper you sent me have to prove 1/x + 1/y + 1/z = 0, not 1/x + 1/y - 1/z = 0. – 2017-01-22
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    I think you have to edit your question here. – 2017-01-22
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Take $$2^x =3^y =6^{-z} =\alpha $$ $$\Rightarrow x=\log_2 \alpha =\frac {\log \alpha}{\log 2}; y =\frac {\log \alpha}{\log 3}; -z=\frac {\log \alpha}{\log 6} $$

Can you take it from here?

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    1/x+1/y-1/z = log(a)/log36 but i dont know log(a)... – 2017-01-22
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    Actually $$\frac {1}{x} +\frac {1}{y}-\frac {1}{z} =\frac {\log 36}{\log \alpha} $$ Anyways there lies the mistake in the options. Clearly the setter intended to find $\sum \frac {1}{x} $ to get an easy answer- $0$ but unfortunately a typo in the question!! – 2017-01-22
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set $$2^x=t,3^y=t,6^{-z}=t$$ then we have $$\ln(6)=-\frac{\ln(t)}{z}$$ and $z \ne 0$ thus we get $$\ln(2)+\ln(3)=-\frac{\ln(t)}{z}$$ now plugging the given Terms in this equation $$\frac{\ln(t)}{x}+\frac{\ln(t)}{y}=-\frac{\ln(t)}{z}$$ can you proceed?

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    1/x+1/y-1/z = log(t)/log36 but i dont know log(t)... – 2017-01-22
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    you can divide by $\ln(t)$ since it is not zero – 2017-01-22
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So $2^x=3^y=6^w$ (where $w=\pm z$ may be confused by a typo). We can sort this without logs etc. So this is to show a method rather than resolve typos.

Then we have $2^{x-w}=3^w$. Raise this to the power $y$ to obtain $2^{y(x-w)}=3^{wy}$. Now substitute $3^y=2^x$ so that $2^{y(x-w)}=2^{wx}$.

We can equate exponents to give $yx-yw=wx$ and divide by $wxy\neq 0$ to get $\frac 1w-\frac 1x=\frac 1y$ or$$\frac 1x+\frac 1y-\frac 1w=0$$

Note that if $wxy=0$ then the sum is undefined, but the original equation is satisfied by $w=x=y=0$, so this should have been excluded as a possibility.