Results 1 to 8 of 8

Math Help - Complicated Logarithms

  1. #1
    PacManFrogg
    Guest

    Complicated Logarithms

    Please help. I need to solve this equation:

    e^x = 5x - 3

    Thank you.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,829
    Thanks
    123
    Quote Originally Posted by PacManFrogg
    Please help. I need to solve this equation:
    e^x = 5x - 3
    Thank you.
    Hello,

    by your equation you calculate the intercepting points of graphs of an exponential function and a line. I've attached a diagram to show the situation.
    I don't know a method to solve this equation algebraically. Instead I would use an iteration(?) (I mean that you have to construct a convergent sequence to obtain the solution). I'll demonstrate a method which I know under the name Newton's method:
    1. Transform the equation into a function: f(x)=e^x-5x+3
    2. Choose a x-value for a start. Here I would say: x_0 = 1
    3. The next x-value is calculated by:

    x_{n+1}=x_n-{f(x_n)\over f'(x_n)} where f'(x) is the first derivation of f. With your problem it is: f'(x) = e^x-5

    So you get:

    x_{1}=1-{{e^1-5+3}\over {e^1-5}}\approx 1.3147...

    Now plug in this value for x_n and you'll get the next x-value which comes closer to the solution:
    x_1 = 1.3147...
    x_2 = 1.4323...
    x_3 = 1.4653...
    x_4 = 1.4687...
    x_5 = 1.4688...
    x_6 = is the same as x_5. That means you substract zero from the previous x-value which is only possible if f(x) = 0. That was intended. So you've found one solution.

    Change the x_0-value to 2 and you'll get the second solution: x = 1.74375...

    Greetings

    EB
    Attached Thumbnails Attached Thumbnails Complicated Logarithms-gerade_exp.gif  
    Last edited by earboth; May 25th 2006 at 01:38 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by earboth
    Hello,

    by your equation you calculate the intercepting points of graphs of an exponential function and a line. I've attached a diagram to show the situation.
    I don't know a method to solve this equation algebraically. Instead I would use an iteration(?) (I mean that you have to construct a convergent serie to obtain the solution). I'll demonstrate a method which I know under the name Newton's method:
    1. Transform the equation into a function: f(x)=e^x-5x+3
    2. Choose a x-value for a start. Here I would say: x_0 = 1
    3. The next x-value is calculated by:

    x_{n+1}=x_n-{f(x)\over f'(x)} where f'(x) is the first derivation of f. With your problem it is: f'(x) = e^x-5

    So you get:

    x_{1}=1-{{e^1-5+3}\over {e^1-5}}\approx 1.3147...

    Now plug in this value for x_n and you'll get the next x-value which comes closer to the solution:
    x_1 = 1.3147...
    x_2 = 1.4323...
    x_3 = 1.4653...
    x_4 = 1.4687...
    x_5 = 1.4688...
    x_6 = is the same as x_5. That means you substract zero from the previous x-value which is only possible if f(x) = 0. That was intended. So you've found one solution.

    Change the x_0-value to 2 and you'll get the second solution: x = 1.74375...

    Greetings

    EB
    There is another solution near x=1.75

    RonL
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,829
    Thanks
    123
    Quote Originally Posted by CaptainBlack
    There is another solution near x=1.75
    RonL
    Hello,

    you are right.
    But as I've mentioned above it is x = 1.74375...

    Greetings

    EB
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by earboth
    Hello,

    you are right.
    But as I've mentioned above it is x = 1.74375...

    Greetings

    EB
    Sorry I missed that.

    The problem can also be solved using the Lambert W function.

    RonL
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,829
    Thanks
    123
    Quote Originally Posted by CaptainBlack
    The problem can also be solved using the Lambert W function.
    RonL

    Hello,

    and äääähem. Well I've heard the name of Heinrich Lambert. But I don't know his famous W-function (I believe to remeber that it deals somehow with complex arguments and differential equations(?)), and worse: I wouldn't be able to explain how to use.

    Greetings

    EB
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by earboth
    Hello,

    and äääähem. Well I've heard the name of Heinrich Lambert. But I don't know his famous W-function (I believe to remeber that it deals somehow with complex arguments and differential equations(?)), and worse: I wouldn't be able to explain how to use.

    Greetings

    EB
    The function, xe^x is bijective for x\geq 0 therefore it has an inverse W(x). That is how the lambert function is defined. (It happens to be non-elementary).
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by ThePerfectHacker
    The function, xe^x is bijective for x\geq 0 therefore it has an inverse W(x). That is how the lambert function is defined. (It happens to be non-elementary).
    Unfortunately in this case the x and W(x) involved are not >0 (worse still the
    x involved is not even >-1/e) so we are in a region where W is multi-valued
    (as we should expect since there are two real roots to this problem).

    My mention of W was a sort of joke, theoretically it can be used to write
    the solution in closed form, practically one is better off using numerical
    methods to solve the problem.

    RonL
    Last edited by CaptainBlack; May 25th 2006 at 10:09 PM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. A Complicated Integral
    Posted in the Calculus Forum
    Replies: 10
    Last Post: January 18th 2011, 12:46 PM
  2. complicated equation
    Posted in the Algebra Forum
    Replies: 3
    Last Post: January 5th 2011, 02:26 PM
  3. very complicated algebra
    Posted in the Algebra Forum
    Replies: 1
    Last Post: July 25th 2009, 07:59 AM
  4. complicated p-series
    Posted in the Calculus Forum
    Replies: 9
    Last Post: March 22nd 2009, 05:04 PM
  5. Complicated Series
    Posted in the Calculus Forum
    Replies: 1
    Last Post: May 16th 2008, 07:18 PM

Search Tags


/mathhelpforum @mathhelpforum