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

**JOhkonut** · August 28th 2008, 06:14 PM

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.

**Jhevon** · August 28th 2008, 06:27 PM

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 $\ln |x| = \left \{ \begin{array}{lr} \ln x & \mbox{ if } x > 0 \\ & \\ \ln (-x) & \mbox{ if } x < 0 \\ & \\ \text{Does not exist} & \mbox{ if } x = 0 \end{array} \right.$

**JOhkonut** · August 28th 2008, 06:38 PM

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...

**Jhevon** · August 28th 2008, 07:14 PM

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

**JOhkonut** · August 28th 2008, 07:19 PM

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.

**Soroban** · August 28th 2008, 07:28 PM

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!
.

**Jhevon** · August 28th 2008, 08:16 PM

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

**Chris L T521** · August 28th 2008, 10:11 PM

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

**mr fantastic** · August 29th 2008, 02:33 AM

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 $g(x)<1 \Leftrightarrow x<a \text{ where } g(a)=1$

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

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 ......

**Jhevon** · August 29th 2008, 06:21 AM

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!

**NonCommAlg** · August 29th 2008, 07:18 AM

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 $\frac{1}{|x|}.$

**Moo** · August 29th 2008, 07:25 AM

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

**Moo** · August 29th 2008, 07:53 AM

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 $g(x)<1 \Leftrightarrow x<a \text{ where } g(a)=1$

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

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

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