Proof that function ln$\frac {(2-e^x)} {(3+2e^x)}$ is surjective...

Question

I would appreciate some help with proving that function ln$\frac {(2-e^x)} {(3+2e^x)}$ is surjective... I get $x = \text{ln}\frac {(2-3y)} {(3y+1)}$, but I don't know how to get from here on. What is the official result? Thank you in advance for answers :)

Answer 1

Let $x \in \mathbb R$. Let $y = e^x$ then $y > 0.$ Then we have

$$ \frac{2-0}{3-2 \cdot 0} = \frac 2 3>\frac{2-y}{3+2y}$$ since the function $t \mapsto \frac{2-t}{3+2t}$ decreases on $t > \frac {-2}{3}.$ Since the logarithm is an increasing function we have

$$ \ln \frac 2 3 > \ln \frac{2-e^x}{3+2e^x}$$

Therefore the function $f :\mathbb R \rightarrow \mathbb R : x \mapsto \ln \frac{2-e^x}{3+2e^x}$ is not surjective since

$$ \text{Im} f \subset (-\infty, \ln \frac 2 3) \neq \mathbb R.$$

Answer 2

$$f(x)=\ln\frac {(2-e^x)} {(3+2e^x)}$$ The domain of $f(x)$ is $(-\infty, \ln 2)$ and the range is $(-\infty, \ln(2/3))$and $$f'(x)=\frac{-7e^x}{(2-e^x)(3+2e^x)}<0$$ which is negative in the domain. So it is monotonically decreasing. Hence $f:(-\infty, \ln 2) \rightarrow (-\infty,\ln(2/3))$ is both Injective (one to one) and Surjective (onto). Hence it is Bijective.