Solve this inequality

Question

We have this inequality :

$1-x^2-\log x \ge 0$

How we can solve this inequality ? (Note : It is natural $\log$)

(In general , I have problem with inequalities involving logarithms and exponents functions.)

Answer 1

Let $x > 1$. Then $1-x^2<0$ and $\log x>0$. In this case the inequality is false.

Let $0<x < 1$. Then $1-x^2 > 0$ and and $\log x < 0$. In this case the inequality is true.

The case $x=1$ should be clear.

Answer 2

If the $\log$ denotes the natural logarithm (i.e., base $e$), then

Hint:

• What is the sign of $\log x$?
• When $\log x$ is negative, what is the sign of the expression?
• What about when $\log x$ is positive?

Answer 3

Notice that we must have $x > 0$ for $\log x$ to be defined. Next, notice that $1 - x^2 - \log x = 0$ if and only if $1 - x^2 \geq \log x$.

The function $f_1(x) = 1 - x^2$ is strictly decreasing for the interval $(0;\infty)$ while $f_2(x) = \log x$ is strictly increasing for the same interval. We have that $\lim_{x \rightarrow 0} f_1(x) = 1$, while $\lim_{x \rightarrow 0} f_2(x) = -\infty$.

The graphs of $f_1$ and $f_2$ intersect at $x=1$, since $1 - 1^2 = \log 1 = 0$. We must therefore have that $0 < x \leq 1$ is the required solution.

Answer 4

if we have $$f(x)=1-x^2-ln(x)$$ then we get $$\lim_{ x \to 0}f(x)=+\infty$$ and $$f'(x)$$ is given by $$f'(x)=-2x-\frac{1}{x}$$ for $$x>0$$ and $$\lim_{x \to \infty}f(x)=-\infty$$ thus we have only one intersection point with the $$x$$ axis whis is given by $x=1$