Uniqueness of solution for transcedental equation on the open set

Question

What is the best way to prove that x=a is the unique solution for the equation $$\frac{2a}{x} = \exp (2-\frac{2x}{a}) +1$$ for $x>0$ ? Intermediate Value Theorem does not work since the interval is open. Thank you for your time.

Answer 1

HINT

Let consider $ay=x>0$

$$\frac{2a}{x} = e^{2-\frac{2x}{a}} +1\iff \frac{2}{y}-1=e^{2-2y}\iff\frac{2-y}{y}=e^{2-2y}\iff\frac{y}{2-y}=e^{2y-2}$$

Let $f(y)=\frac{y}{2-y}$ and $g(y)=e^{2y-2}$ and observe that

• $f(y)> 0$ for $0<y<2$ and $f(y)<0$ for $y>2$ but $g(y)>0$ thus consider $0<y<2$

• $f(0)=0$ and $g(0)>0$

• $\lim_y\to2^- f(y)=+\infty$ and $g(2)=1$ thus by IVT a solution exists for $0<y<2$

• for unicity use monoticity