Find all numbers $a$ for which the equation $a3^x+3^{-x}=3$ has a unique solution

Question

Find all numbers $a$ for which the equation $a3^x+3^{-x}=3$ has a unique solution $x$.

How should I approach this? I don't really see any good way to start the problem, I've seen a similar problem before and If my memory serves me right they used the discriminant somehow(?)

Answer 1

Multiply everywhere by $3^x$ to get $$ a(3^x)^2+1=3\cdot 3^x $$ With substitution $y=3^x$ and a little rearranging, we get $$ ay^2-3y+1=0 $$ We are looking for the values of $a$ which gives us a single positive solution. The discriminant helps here, but it won't be the complete solution.

Answer 2

$$a\cdot 3^x + \frac{1}{3^x}=3$$ $$\frac{a\cdot 3^{2x} - 3\cdot 3^x + 1}{3^x} = 0$$ $$a\cdot 3^{2x} - 3\cdot 3^x + 1 = 0$$

1. So we have a quadratic equation. If we want a unique solution, the discriminant ($D$) must be equal to 0: $$D = 9 - 4a = 0$$ $$a = \frac{9}{4}$$
2. But this is not the end. The equation has another unique solution if $a \leq 0$. If $a \leq 0$, then $D \geq 9$, and we have $$y_1 = \frac{3 - \sqrt{D}}{2a}$$ $$y_2 = \frac{3 + \sqrt{D}}{2a} $$ so $y_2$ is not a solution (because $3^x$ can't be negative) and we have a unique solution in this case too. $$Answer: a \in (-\infty , 0] \cup \left\{\frac{9}{4}\right\}$$

Answer 3

We can replace $3^{-x}$ by $t>0$ and the equation reads

$$\frac at+t=3.$$

From this we draw

$$a=t(3-t)$$ which describes a downward parabola through $(0,0)$ and $(3,0)$. This equation has a single positive solution in $t$ when $\color{green}{a<0}$, or a double root at the vertex, $\color{green}{a=\left(\dfrac32\right)^2}$ (indeed $\dfrac32>0$).

Answer 4

We have $a=3^{1-x}-3^{-2x}.$

Since $$\left(3^{1-x}-3^{-2x}\right)'=\frac{(2-3^{x+1})\ln3}{3^{2x}},$$ we see that for $x=\log_32-1$ the expression $3^{1-x}-3^{-2x}$ gets a maximal value $\frac{9}{4}.$

Thus, for $a=\frac{9}{4}$ our equation has an unique root.

Also, since $$\lim_{x\rightarrow+\infty}\left(3^{1-x}-3^{-2x}\right)=0$$ and $$\lim_{x\rightarrow-\infty}\left(3^{1-x}-3^{-2x}\right)=-\infty,$$we see that any $a\leq0$ is also valid.

Another way.

For $a=0$ our equation has an unique real root.

Let $a<0$.

Thus, $$a3^{2x}-3\cdot3^x+1=0,$$ which is a quadratic equation of $3^x$ with negative product of roots and since $3^x>0$, we obtain that for any $a<0$ our equation has an unique real root.

If $a>0$, so the discriminant should be non-negative, which gives $a\leq\frac{9}{4},$ which says that for any $0<a<\frac{9}{4}$ our equation has two different real roots.

For $a=\frac{9}{4}$ we obtain that our equation has an unique real root.