# Solve e^x-lnx<=e/x ?

## Solve for $x$, ${e}^{x} - \ln x \le \frac{e}{x}$

May 19, 2018

so the solution of this inequality make it true $x \in \left(0.1\right]$

#### Explanation:

consider $f \left(x\right) = {e}^{x} - \ln x - \frac{e}{x}$ ,we have

$f ' \left(x\right) = {e}^{x} - \frac{1}{x} + \frac{e}{x} ^ 2$

argue that $f ' \left(x\right) > 0$ for all real x and conclude noting that $f \left(1\right) = 0$

$f \left(1\right) = e - \ln 1 - e = 0$

consider the limit of f as x goes to 0

${\lim}_{x \rightarrow 0} {e}^{x} - \ln x - \frac{e}{x}$

${\lim}_{x \rightarrow {0}^{+}} {e}^{x} - \ln x - \frac{e}{x} = - \infty$

In other words, by showing $f ' \left(x\right) > 0$ you show that the function is strictly increasing, and if $f \left(1\right) = 0$ that means that $f \left(x\right) < 0$
for $x < 1$ because the function always grows.

from the definition of $\ln x$

$\ln x$ is defined for each $x > 0$

from the definition of ${e}^{x}$

${e}^{x}$ is defined for each $x \ge 0$

but $\frac{e}{x} = \frac{e}{0}$ undefined

so the solution of this inequality make it true $x \in \left(0.1\right]$