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If $$2^x=3^y=6^{-z}$$ and $x,y,z \neq 0 $ then prove that:$$ \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0$$

I have tried starting with taking logartithms, but that gives just some more equations.

Any specific way to solve these type of problems?

Any help will be appreciated.

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$2^x=3^y=6^{-z}=k $ say, then $2= k^{1\over x},3=k^{1\over y},6=k^{-1\over z}$ now can you go on?

then $k^{-1\over z}=6=2\times 3 = k^{1\over x}\times k^{1\over y}=k^{{1\over x}+{1\over y}}$

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

so$$x = \log_2k$$ $$ y = \log_3k$$ $$z= -\log_6k$$

so $$\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \log_k2 + log_k3 -\log_k6$$ $$=\log_k{\frac{2\times3}{6}}$$ $$=0$$

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You have not dealt with the possibility $k=1$ – Mark Bennet Aug 31 '13 at 17:04
@MarkBennet k= 1 is not possible because argument1 : then $2^x = 1 $ => $2^x = 1^x $ which implies 2=1 which is a false statement . argument2: $2^x = 1 $ means $2^x = 2^0$ so x=0, x=0 makes 1/x infinity so the equation will not be valid for x=0 , so this problem becomes invalid so assumed $2^x \neq 1$ – Harish Kayarohanam Aug 31 '13 at 17:40
@MarkBennet Now the question after someone edited, excludes condition for x,y,z$\neq 0$ – Harish Kayarohanam Aug 31 '13 at 17:47
Thanks for letting me know. However, in the original equation it is possible for $k$ to be equal to $1$ unless the value is excluded we have $0=-0$ so that $1=2^0=3^0=6^{-0}$ and taking logarithms just gives $0=0=0$. I found it interesting that most of the solutions failed to notice this possibility as a result of an unusually hidden division by zero. Take care of special cases. – Mark Bennet Aug 31 '13 at 17:55
Ya . Good observation. Thanks .Good learning from you. – Harish Kayarohanam Aug 31 '13 at 18:02

Take logarithms (any base) to obtain $$x\log 2= y\log 3=-z\log 6=-z\log 3-z\log 2$$

Then note that $(y+z)\log 3=-z\log 2$ and $x\log 2 = y\log 3$

Multiply left- and right- hand sides to obtain $$(xy+yz)\log 3 \log 2=-yz\log 2\log 3$$

Whence $xy+yz+zx=0$

Note that we have done no division so far, except by the non-zero term $\log 2\log 3$, and also that $x=y=z=0$ is a solution. If $xyz\neq 0$ we can divide by $xyz$ to obtain the required equation.

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Note that it is trivial that if any of the exponents is zero, they all are. – Mark Bennet Aug 31 '13 at 16:53
Note that the condition equivalent to $xyz\neq 0$ in the question was added later. I leave this solution here, because the other solutions proposed simply don't engage with the fact that this condition is significant - it isn't obvious how they use it. This method does, and deliberately so. – Mark Bennet Aug 31 '13 at 20:38

As $\log_aa=1$ and $\log_a(bc)=\log_ab+\log_ac,$

applying logarithm wrt $2,$


$x=y\log_23\implies \log_23=\frac xy$

and put this value of $\log_23$ in $\displaystyle x=-z(1+\log_23)$

to eliminate $\log_23$ and simplify.

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Please disclose the mistake that has caused the down-vote. If its logarithm, then some other answers also have used logarithm – lab bhattacharjee Aug 31 '13 at 17:00
I thought this illustrated a useful property of logarithms - and solved the problem. – Mark Bennet Aug 31 '13 at 17:07
@MarkBennet, thanks for your feedback. Wish there was a provision to pinpoint mistake(unless obvious) before downvoting – lab bhattacharjee Aug 31 '13 at 17:42

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