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$$ x \in \mathbb R $$

$$\frac{9^{x}-9^{-x}}{3^{x}+3^{-x}} =-80\cdot 3^x$$

How one should go about solving this? I tried to solve it on my own but I ended up with : $${3^{2x+1}+1 = -80\cdot 3^x}$$

What exactly am I doing wrong here? I tried a few times but wasn't able to solve it. I would really appreciate if someone took a shot at this for me.

Thanks in advance!

The answer should be = $-2$

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  • $\begingroup$ I suppose the right-hand side should be $-80\cdot 3^x$ rather than $-80.3^x$. Otherwise the proposed answer is false. Also, a hint: $9^{x} = (3^x)^2$ and $a^2-b^2 = (a+b)(a-b)$. $\endgroup$ – TZakrevskiy Nov 6 '17 at 15:15
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Let $3^x=t$.

$$\frac{t^2-(t^{-1})^2}{t+t^{-1}}=-80t$$ or

$$t-t^{-1}=-80t$$

or

$$81t^2=1.$$

Then obviously, $x=-2$ is the only solution.

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Hint:

$$9^x-9^{-x}=80-80\cdot9^x$$

Set $9^x=a$ to find $$81a^2-80a-1=0$$

For real $x,a>0$

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