$p3^x+2\cdot 3^{-x}=1$
I got this down to a quadratic equation by marking $3^x$ as $t$ and I fiddled with the stuff and got some solutions that apparently don't fit the real one in the textbook was.
I'm not even sure what "unique" solution means in this case. I thought that it meant all p for which the solutions of the quadratic equation are of different signs, but solving for that didn't give me the right answer.
How do I approach this?
To have any solution you need the quadratic equation
$$p y^2+2=y$$
to have a positive solution, since $y=3^x$ is always positive. To have a unique solution you need there to be only one positive solution. This will happen when either there is only one solution which happens to be positive, or when there is a positive and a nonpositive solution. In standard form you have
$$py^2-y+2=0.$$
So the discriminant is $1-8p$. There is only one solution if this is zero, and it is positive if $p<0$. There are two solutions if this is positive, and they have different signs if $p<0$ (since then $b^2-4ac>b^2$, so $\sqrt{b^2-4ac}>|b|$).
Can you finish from here?