How to solve exponential inequality with $x$

Question

I need to solve the following inequality.

$$\ln(x) - x > 0.$$

I oddly remember that it can only be done by using the graph... Is it true?

I have the same problem with

$$e^x(x-1)>-2.$$

Thanks!

Answer 1

I'll only address the first one. We essentially need to find the values of $x$ such that the function $\ln x - x$ is positive. A rough estimate can be given by graphing it, and that is a good place to start. In the first case, graphing it reveals that there are no solutions. Though the graph is positive up until $-1$, in general logarithms are not defined for negative numbers. The rest of the graph is negative, so there are no solutions.

For the next inequality, I would recommend that you add $2$ to both sides, and inspect the graph of $e^x(x-1)+2$ to see where it's greater than zero.

Answer 2

Let $f(x)=\log(x)-x$. Note that $\log(x)$ is meaningful for $x>0$. Then, $f'(x)=\frac{1}{x}-1$ so that for $0<x<1$, the function is increasing and for $x>1$, the function is decreasing. At $x=1$, we have $f(1)=0-1=-1$. Together, these last 2 sentences say $f$ is always less than or equal to $-1$. In other words, there is no real $x$ such that $\log(x)-x>0$.

Similarly, let $g(x)=e^x(x-1)$. Then $g'(x)=xe^x$ so that $g$ decreases for negative $x$ and increases for positive $x$. When $x=0$, $g$ evaluates to $1(0-1)=-1$. This means $g$ is always greater than or equal to $-1$. In particular, for all real $x$, $e^x(x-1)>-2$.

Answer 3

These are both consequences of the standard inequality $$ 1+x\le e^x \tag1 $$ How to best prove (1) depends on how you develop the theory of the exponential function. If you know that $e^x$ is its own derivative, then you can argue that $$ e^x - 1 = \int_0^x e^t\,dt \ge \int_0^x 1\,dt = x $$ (because $e^t\ge 1$ if $t\ge 0$ and $e^t\le 1$ if $t\le 0$; note that we could have $x<0$ so that the integration interval is backwards).

Anyway, once you have (1), take logs to get $$ \ln(1+x) \le x $$ and then replace $x$ with $x-1$ to get $$ \ln(x)\le x-1 $$ which implies your first inequality.

For the second, if $x\ge 1$ then multiplying (1) by $x-1$ yields $$ e^x(x-1) \ge (1+x)(x-1) = x^2-1 \ge -1 $$ which implies your second inequality in the case $x\ge 1$. On the other hand, if $x<1$ then replacing $x$ with $-x$ in (1) yields $$ e^{-x} \ge 1-x $$ taking reciprocals (which reverses the inequality because both sides are positive in this case) yields $$ e^x \le \frac1{1-x} $$ Multiplying by $x-1$ (which is negative) yields $$ e^x(x-1) \ge -1 $$ which implies your second inequality in the case $x<1$.