Dealing with Powers of Three!

$\ Satyajit$ I think I managed to find a much easier way to solve the problem without using logs. Firstly multiply the equation by $\ 3^x$ then rearrange to obtain: $\ n . 3^{2x} -3.3^x +1 = 0$ Now let $\ y = 3^x$ : $\ ny^2 -3y +1 = 0$ To obtain a unique solution for $\ x$ we need a corresponding unique solution for $\ y$ so the discriminant must equal zero: $\ b^2 - 4ac = 9 - 4n = 0 \therefore\ n = \frac{9}{4} = 2.25$

We will prove that the required numbers ' $n$ ' are those which satisfy $n \leq 0$ or $n = \dfrac94$ . Let $n,x$ be real numbers. Let $f(x) = 3^{1-x} - 3^{-2x}$ .

Then $n \cdot 3^x + 3^{-x} = 3 \Rightarrow f(x) = n$ .

For all real numbers $x$ : $f'(x) = \ln(3) \cdot 3^{-2x} \left( 2 - 3^{1+x} \right)$ .

Let $x_0 = \dfrac{\ln(\frac23)}{\ln(3)}$ (that is $3^{1+x_0} = 2$ ). We then have:

and since $3^{1+x_0} = 2$ , we have $f(x_0) = 3 \cdot \dfrac32 - \left( \dfrac32 \right)^2 = \dfrac94$ .

From this, we can easily deduce that the equation $f(x) = n$ has a unique solution if and only if $n = f(x_0) = \dfrac94 \text{ or } n \leq 0$ . But since we need positive $n$ , we end up with $n = \dfrac94 = \boxed{2.25}$ .