For how many real numbers x does e^x=ln|x|? - Page 2

NonCommAlg · #11 $Old$ 08-29-2008, 09:18 AM

Quote:

Originally Posted by Moo $View Post$

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 · #12 $Old$ 08-29-2008, 09:25 AM

Quote:

Originally Posted by NonCommAlg $View Post$

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 · #13 $Old$ 08-29-2008, 09:53 AM

Hello !

Quote:

Originally Posted by JOhkonut $View Post$

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