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    Default Interesting equation

    e^x*lnx = 1

    I've been trying to solve that but haven't succeeded.
    Anyone wanna give it a shot?

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    So this is the same as

    ln(x)*e^x = 1 right?
    That means ln(x)*e^x - 1 = 0

    Put that in my TI-84, and the answer comes out to be x = 1.310

    I'm not sure how to do it by hand, I usually do hard equation solving with my calculator.

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    Quote Originally Posted by 3Garrett3 View Post
    So this is the same as

    ln(x)*e^x = 1 right?
    That means ln(x)*e^x - 1 = 0

    Put that in my TI-84, and the answer comes out to be x = 1.310

    I'm not sure how to do it by hand, I usually do hard equation solving with my calculator.
    Hehe, yeah I know how to approximate roots but I'd like to know the exact answer.

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    Quote Originally Posted by 3Garrett3 View Post
    So this is the same as

    ln(x)*e^x = 1 right?
    That means ln(x)*e^x - 1 = 0

    Put that in my TI-84, and the answer comes out to be x = 1.310

    I'm not sure how to do it by hand, I usually do hard equation solving with my calculator.
    why do you have to subtract the one to use it in the calculator? you shouldnt have to do that

    and isnt ln(x) the reverse of e^x?
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    Quote Originally Posted by Awkwardsaw View Post
    why do you have to subtract the one to use it in the calculator? you shouldnt have to do that

    and isnt ln(x) the reverse of e^x?
    Well, my calculator is the best that IB allows, and they don't allow something where you can just enter the equation into it and it tells you the exact answers and how to get it. I have a *Solver* function, which must be equal to 0 in order for it to be used, and it only gives approximate answers.

    And ln(x) is the log(base e)x. I know that ln(e^x)) is simply x. But when they are multiplied together, I don't think the same rules apply.

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    Quote Originally Posted by 3Garrett3 View Post
    Well, my calculator is the best that IB allows, and they don't allow something where you can just enter the equation into it and it tells you the exact answers and how to get it. I have a *Solver* function, which must be equal to 0 in order for it to be used, and it only gives approximate answers.

    And ln(x) is the log(base e)x. I know that ln(e^x)) is simply x. But when they are multiplied together, I don't think the same rules apply.
    *cough*get a casio*cough*

    and kay, i havnt used logs for ever lol. but it seems like there should be somesort of cancelation(sp)?
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    Quote Originally Posted by Awkwardsaw View Post
    *cough*get a casio*cough*

    and kay, i havnt used logs for ever lol. but it seems like there should be somesort of cancelation(sp)?
    I was provided with a free TI-84 Plus just for doing IB. I wouldn't spend money on a calculator which either a) wasn't allowed by IB or b)could be upgraded when I got to uni.

    I'm not sure yet what kind of calculator I'll be allowed in my engineering courses next year, so I wont be getting one until then.

    But ontopic, my least favourite portion of the math curriculum was the logs and e^x, so I may have forgotten a rule about cancellations, but I don't remember there being one when they were multiplied.

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    Quote Originally Posted by 3Garrett3 View Post
    I was provided with a free TI-84 Plus just for doing IB. I wouldn't spend money on a calculator which either a) wasn't allowed by IB or b)could be upgraded when I got to uni.

    I'm not sure yet what kind of calculator I'll be allowed in my engineering courses next year, so I wont be getting one until then.

    But ontopic, my least favourite portion of the math curriculum was the logs and e^x, so I may have forgotten a rule about cancellations, but I don't remember there being one when they were multiplied.
    e^lnx = x

    But if you like Texas Instruments like I do, get a TI-89, it's pure awesomeness!
    http://education.ti.com/educationpor...us_ti89ti.html

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    Quote Originally Posted by marpis View Post
    e^lnx = x

    But if you like Texas Instruments like I do, get a TI-89, it's pure awesomeness!
    http://education.ti.com/educationpor...us_ti89ti.html
    I do like them, but I am not sure what my Uni is going to say, until I get there next year. So I'll be safe rather than sorry.

    e^lnx = x as well as ln(e^x) = x correct?

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    Indeed interesting equation.. I've failed to solve it .. Curious to see the answer!
    Verrekte Koekwous

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    Quote Originally Posted by 3Garrett3 View Post
    I do like them, but I am not sure what my Uni is going to say, until I get there next year. So I'll be safe rather than sorry.

    e^lnx = x as well as ln(e^x) = x correct?
    yep.

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    is x ln x power or its (e^x)(lnX)


    lol here is my poor attempt
    Last edited by Main; 05-10-2010 at 05:04 AM.
    Oh Hai Dar

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    line 8??? log x = 10^x ????
    Infractions, reputation, reflection, the dark side of scripting, they are.

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    yeah e^x cancels out.

    Besides my answer was obviously wrong, but a attempt wasn't bad .
    Oh Hai Dar

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    I mean... at line 8 you rewrite log x as 10^x.... which isn't even true (e.g. x =1 ==> 0 = 10)
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    Yeah, I did them while half a sleep lol.
    I showed my teacher this, she couldn't do it. I will attempt it again.

    Zomg... just use netwon's method
    Last edited by Main; 05-11-2010 at 01:54 AM.
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    Quote Originally Posted by Main_Ftw View Post
    Zomg... just use netwon's method
    If you want an approximation zomg just use texas instrument

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    Exact answer:
    http://mathbin.net/46986

    Following the latter derivation, you can use the alternating series approximation theorem to find how ever many decimal places you would like!

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    Quote Originally Posted by Nava2 View Post
    Exact answer:
    http://mathbin.net/46986

    Following the latter derivation, you can use the alternating series approximation theorem to find how ever many decimal places you would like!

    Oh, but I see 2 variables there. That's not an exact answer, nor an answer at all, just a different way to express the function.
    And, before anyone dares to say it has no exact answer, here's a proof there is one:
    ln(x) is continous and deriving ]0, infinity[
    e^x is continuous and deriving ]-inf, inf[
    -> ln(x)*e^x is continous and deriving ]0, infinity[
    thus lnx*e^x = 1 has a solution (only one in this case, derivative is always positive)
    Last edited by marpis; 05-11-2010 at 04:42 PM.

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    Quote Originally Posted by marpis View Post
    Oh, but I see 2 variables there. That's not an exact answer, nor an answer at all, just a different way to express the function.
    Actually it is an answer ! Use the theorem I mentioned to work out an approximation, my answer is exact. There is no better way to represent it, nor way to calculate it.

    In fact, the calculators you guys used probably used that infinite sum, and the theorem to calculate it. Modified for speed of course.
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    Quote Originally Posted by Nava2 View Post
    Actually it is an answer ! Use the theorem I mentioned to work out an approximation, my answer is exact. There is no better way to represent it, nor way to calculate it.

    In fact, the calculators you guys used probably used that infinite sum, and the theorem to calculate it. Modified for speed of course.
    You see, your sum has both "n" and "x" there, but only giving values to "n".
    How can you say it gives a number as an answer o_O Where does the x go?

    And I don't want an approximation I want an answer like
    "e^x*lnx = 1 when x = e^2*ln(2)" or something like that.

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    Quote Originally Posted by marpis View Post
    You see, your sum has both "n" and "x" there, but only giving values to "n".
    How can you say it gives a number as an answer o_O Where does the x go?

    And I don't want an approximation I want an answer like
    "e^x*lnx = 1 when x = e^2*ln(2)" or something like that.
    If you would read what I said, you would see there is an exact answer as a sum of fractions.
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    Quote Originally Posted by Nava2 View Post
    Exact answer:
    http://mathbin.net/46986
    Following the latter derivation, you can use the alternating series approximation theorem to find how ever many decimal places you would like!
    ^ There's just the infinite sum, which is not an answer, but another way to express the function.

    Quote Originally Posted by Nava2 View Post
    Actually it is an answer ! Use the theorem I mentioned to work out an approximation, my answer is exact. There is no better way to represent it, nor way to calculate it.

    In fact, the calculators you guys used probably used that infinite sum, and the theorem to calculate it. Modified for speed of course.
    ^ We used the answer to calculate the answer? You only posted a way to find an approximation

    Quote Originally Posted by Nava2 View Post
    If you would read what I said, you would see there is an exact answer as a sum of fractions.
    ^ Sorry but I don't understand.

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    You see the difference between the series you presented as a solution to the function e^x*lnx and these?

    The first one has 2 variables: n, which is given values, but then there's x which gets no values.
    In the wikipedia article you posted there is only one variable in the function, n or k in some.

    It's a function if it has variables that are not given any values.
    It's a number if all variables get known values.

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