find the solution of this ln equation:

ax=lnx

i tried:

what to do next??

i thought of building a taylor series around 0 for ln

but ln(0) is undefined

??

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- February 10th 2009, 10:03 PMtransgalacticfind the number of solution of this ln equation
find the solution of this ln equation:

ax=lnx

i tried:

what to do next??

i thought of building a taylor series around 0 for ln

but ln(0) is undefined

?? - February 11th 2009, 05:58 AMJester
- February 11th 2009, 01:22 PMtransgalactic
i need a mathematical solution

a graph wont help.

in whAT WAY SHOULD I try to prove it? - February 11th 2009, 05:56 PMJester
- February 13th 2009, 09:33 AMtransgalactic
i need to find how many solutions??

- February 13th 2009, 10:00 AMJester
- February 13th 2009, 10:58 AMtransgalactic
if i will find the extreme point

f'(x)=0

then

i need to put points and

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) - February 13th 2009, 11:26 AMJester
You're right. The max is located at and so the actual maximum value is

Now clearly if then and since f is continuous on with f approaching negative infinity as then there are x values say where where and so by the intermediate value theorem there exist where giving two solutions to your problem. Similary for and - February 13th 2009, 12:20 PMtransgalactic
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?? - February 13th 2009, 12:27 PMJester
- February 13th 2009, 12:33 PMtransgalactic
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 - February 13th 2009, 12:36 PMtransgalactic
ahh the laws of maximum is for f'(x) not f(x)

i will try to understand this.. - February 13th 2009, 01:13 PMtransgalactic
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

?? - February 13th 2009, 01:35 PMtransgalactic
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 - February 13th 2009, 03:13 PMJester