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Im trying to solve the equation $$3\cdot2^{-2/x} + 2\cdot9 ^{-1/x} = 5\cdot6^{-1/x }$$ So far I tried applying logaritmas but it didnt prove helpful...are there any other ways?

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  • $\begingroup$ First step: replace $1/x$ with a new variable $u$ and solve for $u$. Second: extract a factor of $3$ from each term. I'm not sure where this will lead you, but at least it's a start. $\endgroup$ – John Hughes Jul 5 '14 at 15:15
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Since $9=3^2$ the equation becomes

$$3\times 2^{-2/x}+2\times 3^{-2/x}=5\times2^{-1/x}3^{-1/x}$$ so let $a=2^{-1/x}$ and $b=3^{-1/x}$ then we have

$$3a(a-b)+2b(b-a)=0\iff(a-b)(3a-2b)=0$$ and since $a\ne b$ then $3a-2b=0$ hence $$\frac a2=\frac b3\iff2^{-1/x-1}=3^{-1/x-1}\iff\left(\frac1x+1\right)(\ln3-\ln2)=0\iff x=-1$$

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  • $\begingroup$ nice approach, many thanks $\endgroup$ – Bak1139 Jul 6 '14 at 11:27
  • $\begingroup$ You're welcome. $\endgroup$ – user63181 Jul 6 '14 at 11:30
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Set $2^{-\frac{1}{x}} = a, \ 3^{-\frac{1}{x}} = b$ so the equation becomes $3 a^2 + 2 b^2 -5 ab = 0$.

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