# Math Help - For how many real numbers x does e^x=ln|x|?

1. ## For how many real numbers x does e^x=ln|x|?

For how many real numbers $x$ does $e^x=\ln|x|$?

0,1,2,3 or infinitely many?

Hmmm....

Oh yea, no graphing calculator for this question.

2. Originally Posted by JOhkonut
For how many real numbers $x$ does $e^x=\ln|x|$?

0,1,2,3 or infinitely many?

Hmmm....

Oh yea, no graphing calculator for this question.
hmmm, the easiest way to see the answer is through graphs. of course, i didn't use a graphing calculator either knowing what these graphs look like in your head makes the answer obvious. the other alternative i can think of is too complicated (using Newton's method and a whole lot of speculation)

the answer is 1 by the way ....note that $\ln |x| = \left \{ \begin{array}{lr} \ln x & \mbox{ if } x > 0 \\ & \\ \ln (-x) & \mbox{ if } x < 0 \\ & \\ \text{Does not exist} & \mbox{ if } x = 0 \end{array} \right.$

3. Well, I am supposed to know what the \ln x and e^x graphs look like, and I get how they intersect once, but I have no clue how to show that...

4. Originally Posted by JOhkonut
Well, I am supposed to know what the \ln x and e^x graphs look like, and I get how they intersect once, but I have no clue how to show that...
well, for positive x-values, e^x is always bigger than ln(x), so there is no intersection there. but for negative x values, e^x keeps decreasing tending to zero, while ln(-x) keeps increasing, tending to infinity. the shapes of the graphs make it obvious they will cross each other only once, as one is strictly increasing and the other is strictly decreasing

see below. of course, you should know how this looks in your head, no graphing tool necessary

5. Originally Posted by Jhevon
well, for positive x-values, e^x is always bigger than ln(x), so there is no intersection there. but for negative x values, e^x keeps decreasing tending to zero, while ln(-x) keeps increasing, tending to infinity. the shapes of the graphs make it obvious they will cross each other only once, as one is strictly increasing and the other is strictly decreasing

see below. of course, you should know how this looks in your head, no graphing tool necessary
Thanks a bunch, I saw that when I used my graphing calculator, and just included a sketch of it.

6. Hello, JOhkonut!

For how many real numbers $x$ does $e^x\:=\:\ln|x|$?
You know that $y \:=\:e^x$ looks like, right?
Code:
                |      *
|
|     *
|    *
|  *
*
*   |
*         |
- - - - - - - + - - - - - - -
|

And you know what $y \:=\:\ln x$ looks like.
. . (It's the inverse of the exponential function.)

Then $y \:=\:\ln|x|$ is the same graph
. . but with a reflection over the y-axis.
Code:
                |
*           |           *
*       |       *
- - - - -*- - + - -*- - - - -
*  |  *
* | *
|
*|*
|

Now place them on the same graph and count the intersections.

Darn . . . too slow again!
.

7. Originally Posted by Soroban
Darn . . . too slow again!
That's because you take the time to do those amazing graphs. i don't know how you do it, it seems really hard to me, i'd get too frustrated trying to line everything up

8. Originally Posted by Jhevon
That's because you take the time to do those amazing graphs. i don't know how you do it, it seems really hard to me, i'd get too frustrated trying to line everything up
Ditto

--Chris

9. Originally Posted by Moo
Hello !

One way to do it, without graph, is to study the function $f(x)=e^x-\ln|x|$. Now, be patient and read carefully

IF X<0

$f~'(x)=e^x-\frac 1x$

$f~'(x)<0 \Leftrightarrow e^x<\frac 1x \implies xe^x>1$

Which is not possible, because $e^x>0$ and $x<0$

Therefore, $f~'(x)>0$ and f is strictly increasing when x<0.

$\lim_{x \to -\infty} f(x)=- \infty$

$\lim_{x \to 0^-} f(x)=+ \infty$ (x=0 is a vertical asymptote)

$\boxed{\text{There is one and only one value of x}<\text{0 such that f(x)=0}}$

-----------------------------------------------

IF X>0

$f~'(x)=e^x-\frac 1x$

$f~'(x)<0 \Leftrightarrow e^x<\frac 1x \Leftrightarrow xe^x<1$

Let $g(x)=xe^x$

$g'(x)=e^x+xe^x > 0 \quad \forall x$

Therefore, g is strictly increasing.

$\lim_{x \to 0^+} g(x)=0$

$\lim_{x \to + \infty} g(x)=+\infty$

We can conclude that there exists a>0 such that $g(x)<1 \Leftrightarrow x

So $f~'(x)<0 \Leftrightarrow g(x)<1 \Leftrightarrow x

The function f is decreasing if x<a and then increasing if x>a. Thus f(a) is a minimum of f(x), for x>0.

Since $g(a)=1$, we have $ae^a=1 \implies \frac 1a=e^a$

$f(a)=e^a-\ln(a)=\frac 1a-\ln(e^{-a})=\frac 1a+a$, which is $>0$ since $a>0$

$\boxed{\implies \forall x>0 ~,~ f(x)\ge f(a)>0 \implies f(x) \neq 0 \quad \forall x>0}$

Sorry for the confusion For some reason, I took at least twice the wrong function, and hence the *woops* mistakes
Life's too short. I think I'll wait for it to come out on DVD ......

10. Originally Posted by mr fantastic
Life's too short. I think I'll wait for it to come out on DVD ......
you are too funny!

11. Originally Posted by Moo
Hello !

One way to do it, without graph, is to study the function $f(x)=e^x-\ln|x|$. Now, be patient and read carefully

$f~'(x)=e^x-\frac 1{|x|}$
just to remind you that the derivative of $\ln|x|$ is $\frac{1}{x}$ not $\frac{1}{|x|}.$

12. Originally Posted by NonCommAlg
just to remind you that the derivative of $\ln|x|$ is $\frac{1}{x}$ not $\frac{1}{|x|}.$
True, there's also a problem with g'(x), I'm fixing it lol

13. Hello !
Originally Posted by JOhkonut
For how many real numbers $x$ does $e^x=\ln|x|$?

0,1,2,3 or infinitely many?

Hmmm....

Oh yea, no graphing calculator for this question.
One way to do it, without graph, is to study the function $f(x)=e^x-\ln|x|$. Now, be patient and read carefully

IF X<0

$f~'(x)=e^x-\frac 1x$

$f~'(x)<0 \Leftrightarrow e^x<\frac 1x \implies xe^x>1$

Which is not possible, because $e^x>0$ and $x<0$

Therefore, $f~'(x)>0$ and f is strictly increasing when x<0.

$\lim_{x \to -\infty} f(x)=- \infty$

$\lim_{x \to 0^-} f(x)=+ \infty$ (x=0 is a vertical asymptote)

$\boxed{\text{There is one and only one value of x}<\text{0 such that f(x)=0}}$

-----------------------------------------------
IF X>0

$f~'(x)=e^x-\frac 1x$

$f~'(x)<0 \Leftrightarrow e^x<\frac 1x \Leftrightarrow xe^x<1$

Let $g(x)=xe^x$

$g'(x)=e^x+xe^x > 0 \quad \forall x$

Therefore, g is strictly increasing.
$\lim_{x \to 0^+} g(x)=0$

$\lim_{x \to + \infty} g(x)=+\infty$

We can conclude that there exists a>0 such that $g(x)<1 \Leftrightarrow x

So $f~'(x)<0 \Leftrightarrow g(x)<1 \Leftrightarrow x

The function f is decreasing if x<a and then increasing if x>a. Thus f(a) is a minimum of f(x), for x>0.

Since $g(a)=1$, we have $ae^a=1 \implies \frac 1a=e^a$
$f(a)=e^a-\ln(a)=\frac 1a-\ln(e^{-a})=\frac 1a+a$, which is $>0$ since $a>0$

$\boxed{\implies \forall x>0 ~,~ f(x)\ge f(a)>0 \implies f(x) \neq 0 \quad \forall x>0}$

Sorry for the confusion For some reason, I took at least twice the wrong function, and hence the *woops* mistakes