# find the number of solution of this ln equation

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• Feb 10th 2009, 09:03 PM
transgalactic
find the number of solution of this ln equation
find the solution of this ln equation:
ax=lnx

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

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

??
• Feb 11th 2009, 04:58 AM
Jester
Quote:

Originally Posted by transgalactic
find the solution of this ln equation:
ax=lnx

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

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

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

when there's only one solution.
• Feb 11th 2009, 12:22 PM
transgalactic
i need a mathematical solution
a graph wont help.
in whAT WAY SHOULD I try to prove it?
• Feb 11th 2009, 04:56 PM
Jester
Quote:

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?
• Feb 13th 2009, 08:33 AM
transgalactic
i need to find how many solutions??
• Feb 13th 2009, 09:00 AM
Jester
Quote:

Originally Posted by transgalactic
i need to find how many solutions??

Why not define $\displaystyle f(x) = \ln x -ax$ which iscontinuous on $\displaystyle (0,\infty)$ with $\displaystyle \lim_{x \to 0^+} f(x) = - \infty$ and $\displaystyle \lim_{x \to \infty} = - \infty$. Then consider $\displaystyle f' = \frac{1}{x} - a$ to find the max. If the max is < 0, the no solutions, = 0, then 1 solution and > 0, then 2 solutions.
• Feb 13th 2009, 09:58 AM
transgalactic
if i will find the extreme point
f'(x)=0
$\displaystyle f'(x)=\frac{1}{x}-a$
$\displaystyle \frac{1}{x}=a$
then
$\displaystyle x=\frac{1}{a}$
i need to put points $\displaystyle x1>\frac{1}{a}$ and $\displaystyle x2<\frac{1}{a}$

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)
• Feb 13th 2009, 10:26 AM
Jester
Quote:

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

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 $\displaystyle x = \frac{1}{a}$ and so the actual maximum value is

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

Now clearly if $\displaystyle a < \frac{1}{e}$ then $\displaystyle f_{\text{max}}> 0$ and since f is continuous on $\displaystyle (0,\infty)$ with f approaching negative infinity as $\displaystyle x \to 0^{+},\; x \to \infty$ then there are x values say $\displaystyle x_0,\; x_1$ where $\displaystyle x_1 < \frac{1}{e} < x_2$ where $\displaystyle f(x_1) < 0,\;f(x_2) < 0$ and so by the intermediate value theorem there exist $\displaystyle c_1, \;c_2,\;c_1 \ne c_2$ where $\displaystyle f(c_1) = 0,\;f(c_2) = 0$ giving two solutions to your problem. Similary for $\displaystyle a = \frac{1}{e}$ and $\displaystyle a > \frac{1}{e}$
• Feb 13th 2009, 11:20 AM
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??
• Feb 13th 2009, 11:27 AM
Jester
Quote:

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 $\displaystyle f(x_1) < 0$ to $\displaystyle f(\frac{1}{a}) > 0$ to $\displaystyle f(x_2) < 0$ there's the two.
• Feb 13th 2009, 11:33 AM
transgalactic
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
• Feb 13th 2009, 11:36 AM
transgalactic
ahh the laws of maximum is for f'(x) not f(x)

i will try to understand this..
• Feb 13th 2009, 12:13 PM
transgalactic
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

??
• Feb 13th 2009, 12:35 PM
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
• Feb 13th 2009, 02:13 PM
Jester
Quote:

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