find the number of solution of this ln equation

**transgalactic** · February 10th 2009, 09:03 PM

find the solution of this ln equation:
ax=lnx

i tried:
$ e^{(ax)}=x $
$ e^{(ax)}-x=0 $

what to do next??
i thought of building a taylor series around 0 for ln
but ln(0) is undefined

??

**Danny** · February 11th 2009, 04:58 AM

Originally Posted by transgalactic

find the solution of this ln equation:
ax=lnx

i tried:
$ e^{(ax)}=x $
$ e^{(ax)}-x=0 $

what to do next??
i thought of building a taylor series around 0 for ln
but ln(0) is undefined

??

Try graphing. There can be either 0, 1 or 2 solutions depending on a. The dividing point is when

$a = \frac{1}{e}$

when there's only one solution.

**transgalactic** · February 11th 2009, 12:22 PM

i need a mathematical solution
a graph wont help.
in whAT WAY SHOULD I try to prove it?

**Danny** · February 11th 2009, 04:56 PM

Originally Posted by transgalactic

i need a mathematical solution
a graph wont help.
in whAT WAY SHOULD I try to prove it?

What are you trying to do? Trying to prove solutions exist (and for what values of a), or find them explicity?

**transgalactic** · February 13th 2009, 08:33 AM

i need to find how many solutions??

**Danny** · February 13th 2009, 09:00 AM

Originally Posted by transgalactic

i need to find how many solutions??

Why not define $f(x) = \ln x -ax$ which iscontinuous on $(0,\infty)$ with $\lim_{x \to 0^+} f(x) = - \infty$ and $\lim_{x \to \infty} = - \infty$ . Then consider $f' = \frac{1}{x} - a$ to find the max. If the max is < 0, the no solutions, = 0, then 1 solution and > 0, then 2 solutions.

**transgalactic** · February 13th 2009, 09:58 AM

if i will find the extreme point
f'(x)=0
$ f'(x)=\frac{1}{x}-a $
$ \frac{1}{x}=a $
then
$ x=\frac{1}{a} $
i need to put points $x1>\frac{1}{a}$ and $x2<\frac{1}{a}$

and if f'(x2)>0 and f'(x1)<0
then its a maximum point

i cant see in your explanation any thing that could tell how to get the number of solution regarding the finding of a maximum point
??
(this function is not a parabola we build a delta >0 in order to deside that there are two solutions)

**Danny** · February 13th 2009, 10:26 AM

Originally Posted by transgalactic

if i will find the extreme point
f'(x)=0
$ f'(x)=\frac{1}{x}-a $
$ \frac{1}{x}=a $
then
$ x=\frac{1}{a} $
i need to put points $x1>\frac{1}{a}$ and $x2<\frac{1}{a}$

and if f'(x2)>0 and f'(x1)<0
then its a maximum point

i cant see in your explanation any thing that could tell how to get the number of solution regarding the finding of a maximum point
??
(this function is not a parabola we build a delta >0 in order to deside that there are two solutions)

You're right. The max is located at $x = \frac{1}{a}$ and so the actual maximum value is

$f_{\text{max}}= f\left( \frac{1}{a}\right) = - \ln a -1$

Now clearly if $a < \frac{1}{e}$ then $f_{\text{max}}> 0$ and since f is continuous on $(0,\infty)$ with f approaching negative infinity as $x \to 0^{+},\; x \to \infty$ then there are x values say $x_0,\; x_1$ where $x_1 < \frac{1}{e} < x_2$ where $f(x_1) < 0,\;f(x_2) < 0$ and so by the intermediate value theorem there exist $c_1, \;c_2,\;c_1 \ne c_2$ where $f(c_1) = 0,\;f(c_2) = 0$ giving two solutions to your problem. Similary for $a = \frac{1}{e}$ and $a > \frac{1}{e}$

**transgalactic** · February 13th 2009, 11:20 AM

the intermediate value theorem says that if
there are two points for which f(x1)<0 and f(x2)>0
then there is a x point between then which is a solution
f(x)=0

i found only one such point in the maximum point

there is no another case where f(x1)<0 and f(x2)>0

from where you got the second solution fact??

**Danny** · February 13th 2009, 11:27 AM

Originally Posted by transgalactic

the intermediate value theorem says that if
there are two points for which f(x1)<0 and f(x2)>0
then there is a x point between then which is a solution
f(x)=0

i found only one such point in the maximum point

there is no another case where f(x1)<0 and f(x2)>0

from where you got the second solution fact??

We can also reverse the signs and the same theorem holds. So from $f(x_1) < 0$ to $f(\frac{1}{a}) > 0$ to $f(x_2) < 0$ there's the two.

**transgalactic** · February 13th 2009, 11:33 AM

but the condition for maximum is

f(x_2) > 0 f(x_1) < 0

i dont have

f(x_2) < 0 f(x_1) < 0

**transgalactic** · February 13th 2009, 11:36 AM

ahh the laws of maximum is for f'(x) not f(x)

i will try to understand this..

**transgalactic** · February 13th 2009, 12:13 PM

so in the maximal point x=1/a

if a<1/e f(x)>0
if a>1/e f(x)<0
so there is one solution..

but i dont get two negative point and a positive in the middle

??

**transgalactic** · February 13th 2009, 12:35 PM

our function is
f(x)=ln(x)-ax
our extreme point is x=1/a
f''(x)=-1/x^2 if f''(x1)>0 X1 is minimum if f''(x2)<0 X2 is maximum
f(1/a)=ln(1/a)-1
we cant know if its minum or maximum because it depends on the value of
parameter a.

we cant assume that our derivative will get a possitive or negative value
when we input to the second derivative a parameter

so i am confused regarding your solution

**Danny** · February 13th 2009, 02:13 PM

Originally Posted by transgalactic

our function is
f(x)=ln(x)-ax
our extreme point is x=1/a
f''(x)=-1/x^2 if f''(x1)>0 X1 is minimum if f''(x2)<0 X2 is maximum
f(1/a)=ln(1/a)-1
we cant know if its minum or maximum because it depends on the value of
parameter a.

we cant assume that our derivative will get a possitive or negative value
when we input to the second derivative a parameter

so i am confused regarding your solution

But $f''(x) = - \frac{1}{x^2} < 0\;\;\forall x > 0$ so it concave down so it's a maximum!

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