How to find $y^{(y^2-6)}$?

Question

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

Answer 1

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.

Answer 2

Note that

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

Answer 3

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

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