Range of a logarithmic function.

Question

Let $a\epsilon \mathbb R_+$ and $b=\exp(-a).$

What is the range of $\{-b^2\log(b)\}$? Does it range $(-\infty,+\infty)$?

How can I show $\frac{-b\log(b)}{1-b}\le 1$ ?

Answer 1

First question: Given that $a>0$

$$-b^2\log(b)=-\left(e^{-a}\right)^2\log(e^{-a})=-\left(e^{-a}\right)^2(-a)=ae^{-2a}$$

therefore you need to find the range of $ae^{-2a}$. Since $a>0$, you can think of this as a positive constant times $e^{-2a}$. This produces the range of $(0,\infty)$.

Second question:

$$\frac{-b\log(b)}{1-b}\le 1 \iff\frac{-e^{-a}\log(e^{-a})}{1-e^{-a}}\le 1 \iff \frac{ae^{-a}}{1-e^{-a}}\le 1 \iff \frac{a}{e^a-1}\le 1$$

which simplifies to

$$a \le e^a -1$$

or $$e^a \ge a+1$$

I would recommend applying Bernoulli's inequality. Alternatively, you could observe that

$$0 < e^a = 1 + a + \frac{a^2}{2} + \frac{a^3}{3!} + \frac{a^4}{4!} + \frac{a^5}{5!} + \cdots$$

Answer 2

b in (0,1); log b in (-$\infty$,0).
$-b^2\log(b)$ in (0,$\infty$).