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Math Help - find the number of solution of this ln equation

  1. #1
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    find the number of solution of this ln equation

    find the solution of this ln equation:
    ax=lnx

    i tried:
    <br />
e^{(ax)}=x<br />
    <br />
e^{(ax)}-x=0<br />

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

    ??
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  2. #2
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    Quote Originally Posted by transgalactic View Post
    find the solution of this ln equation:
    ax=lnx

    i tried:
    <br />
e^{(ax)}=x<br />
    <br />
e^{(ax)}-x=0<br />

    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.
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  3. #3
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    i need a mathematical solution
    a graph wont help.
    in whAT WAY SHOULD I try to prove it?
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  4. #4
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    Quote Originally Posted by transgalactic View Post
    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?
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  5. #5
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    i need to find how many solutions??
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  6. #6
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    Quote Originally Posted by transgalactic View Post
    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.
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  7. #7
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    if i will find the extreme point
    f'(x)=0
    <br />
f'(x)=\frac{1}{x}-a<br />
    <br />
\frac{1}{x}=a<br />
    then
    <br />
x=\frac{1}{a}<br />
    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)
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  8. #8
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    Quote Originally Posted by transgalactic View Post
    if i will find the extreme point
    f'(x)=0
    <br />
f'(x)=\frac{1}{x}-a<br />
    <br />
\frac{1}{x}=a<br />
    then
    <br />
x=\frac{1}{a}<br />
    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}
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  9. #9
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    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??
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  10. #10
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    Quote Originally Posted by transgalactic View Post
    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.
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  11. #11
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    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
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  12. #12
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    ahh the laws of maximum is for f'(x) not f(x)

    i will try to understand this..
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  13. #13
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    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

    ??
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  14. #14
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    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
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  15. #15
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    Quote Originally Posted by transgalactic View Post
    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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