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$$\frac{3}{1-3^{x-2}} + \frac{3}{1-3^{2-x}} = y$$ $$x≠2$$ $$y^{(y^2-6)} = ?$$What is $y^{(y^2-6)}$? Could you please explain to me how to solve this question step by step?

chsdwn
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    $$\dfrac3{1-a}+\dfrac3{1-a^{-1}}=\dfrac{3(1-a)}{1-a}=?$$ for $a\ne1$ – lab bhattacharjee Apr 18 '18 at 10:28
  • thank you lab bhattacharjee. i've solved the question. $y^{(y^2-6)} = 3^3$ – chsdwn Apr 18 '18 at 10:36
  • @RubyMell Please remember that you can choose an answer among the given if the OP is solved, more details here https://meta.stackexchange.com/questions/5234/how-does-accepting-an-answer-work – user Apr 19 '18 at 19:40

3 Answers3

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You have

$$\begin{align} y & = \frac{3}{1-3^{x-2}} + \frac{3}{1-3^{2-x}} \\ & = \frac{3(1-3^{2-x}) + 3(1-3^{x-2})}{(1-3^{x-2})(1-3^{2-x})} \\ & = \frac{3(1-3^{2-x} + 1-3^{x-2})}{(1 - 3^{2-x}-3^{x-2}+3^{x-2+2-x})} \\ & = \frac{3(2-3^{2-x} - 3^{x-2})}{(1 - 3^{2-x}-3^{x-2}+3^{0})} \\ & = \frac{3(2-3^{2-x} - 3^{x-2})}{(2 - 3^{2-x}-3^{x-2})} \\ & = 3 & \text{since } x \not =2\\ \end{align}$$

So

$$y^{y^2-6} = 3^{3^2-6}=3^{9-6}=3^3=27$$

And you're done.

Alex Lostado
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Note that

$$y=\frac{3}{1-3^{x-2}} + \frac{3}{1-3^{2-x}} = \frac{3}{1-3^{x-2}} + \frac{3}{1-\frac1{3^{x-2}}}=\\= \frac{3}{1-3^{x-2}} - \frac{3^{x-1} }{1-3^{x-2}}=\frac{3-3^{x-1}}{1-3^{x-2}}=3\frac{1-3^{x-2}}{1-3^{x-2}}=3 $$

user
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Calling $z = 3^{x-2}$ we have

$$ 3\left(\frac{1}{1-z}+\frac{1}{1-z^{-1}}\right) = y \Rightarrow y = 3 $$

so finally

$$y^{y^2-6} = 3^{9-6} = 27$$

Cesareo
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