How to solve exponential inequality with x 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!

Dean Summers

Dean Summers

Answered question

2022-07-20

How to solve exponential inequality with x
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 & Explanation

Danica Ray

Danica Ray

Beginner2022-07-21Added 15 answers

Let f ( x ) = log ( x ) x. Note that log ( x ) is meaningful for x > 0. Then, f ( x ) = 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 ) = x e 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
Hayley Bernard

Hayley Bernard

Beginner2022-07-22Added 5 answers

These are both consequences of the standard inequality
(1) 1 + x e x
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 = 0 x e t d t 0 x 1 d t = x
(because e t 1 if t 0 and e t 1 if t 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 ) x
ln ( 1 + x ) x
and then replace x with x 1 to get
ln ( x ) x 1
which implies your first inequality.
For the second, if x 1 then multiplying (1) by x 1 yields
e x ( x 1 ) ( 1 + x ) ( x 1 ) = x 2 1 1
which implies your second inequality in the case x 1. On the other hand, if x < 1 then replacing x with x in (1) yields
e x 1 x
taking reciprocals (which reverses the inequality because both sides are positive in this case) yields
e x 1 1 x
Multiplying by x 1 (which is negative) yields
e x ( x 1 ) 1
which implies your second inequality in the case x < 1

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