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Natural log

  1. Jan 27, 2005 #1
    I'm trying to solve this equation:

    2^1-2x = 3 e^x+1

    Using natural logs I got

    x= ln(2/3) -1/ 2ln(2/3) +1

    I plugged this into my solver but I'm not getting the correct answer. Anyone see where I made a mistake?
     
  2. jcsd
  3. Jan 27, 2005 #2

    Hurkyl

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    I think you forgot some parentheses: 2^1-2x means [itex]2^1 - 2x[/itex], not [itex]2^{1 - 2x}[/itex].

    Well, show us all of your work, and maybe we can see what you did wrong. (Yes, I'm including you in that "we")
     
  4. Jan 27, 2005 #3
    ...

    No, the problems reads exactly like that. It says to solve algebraically.
    The problem reads: 2^1-2x=3 e^x+1

    So, I have (2/3)^1-2x=e^x+1

    ln(2/3)^1-2x = ln(e^x+1)

    ln(2/3)(1-2x)= x+1

    ln(2/3) - 2ln(2/3)x = x+1

    x+1= ln(2/3)- 2 ln(2/3)x

    2 ln(2/3)x + x= ln(2/3)-1

    (2 ln(2/3)+1)x = ln(2/3)-1

    x= ln (2/3)-1 / 2 ln(2/3)+1

    This is my solution, but it doesn't check. What do ya think?!?
     
  5. Jan 27, 2005 #4
    [tex]2^1-2x=3 e^x+1[/tex]
    this is not solvable....
     
  6. Jan 27, 2005 #5

    dextercioby

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    ...Exactly. :tongue2: It is not solvable exactly...Write those constants in "human" form and then intersect graphics (an exponential and a line)...

    Daniel.
     
  7. Jan 28, 2005 #6
    Sorry

    Sorry guys! Actually, the exponent is 1-2x for 2 and x+1 for e. I guess I should have been more clear. I hope this make more sense now.
     
  8. Jan 28, 2005 #7
    so this is
    [tex] 2^{1-2x} = 3 e^{1+x} [/tex]

    your first step is wrong already.... i don't even border to read the rest
     
  9. Jan 28, 2005 #8
    Gee, Thanks!

    Well, obviously I know that something is wrong with it, why else would I be asking for help? Anyone have anything USEFUL to add?
     
  10. Jan 28, 2005 #9
    don't divided by the 3... take the log at the very begining...
    ln(a*b) = ln(a) + ln(b)
    the above formulas might be useful for you in your first step RHS....
     
  11. Jan 29, 2005 #10

    HallsofIvy

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    1. Your error was converting [itex]\frac{2^{1-2x}{3}[/itex] to [itex]\(\frac{2}{3}\)^{1=2x}[/itex]. They are not at all the same thing: in the correct formula the denominator is NOT to the 1-2x power.

    2. The very first response noted that you PROBABLY mean [itex]2^{1-2x}[/itex] rather than [itex]2^1- 2x[/itex] which was what you had written. YOUR response to that was "No, the problems reads exactly like that. It says to solve algebraically.
    The problem reads: 2^1-2x=3 e^x+1". I GUARENTEE the problem did NOT "read exactly like that"- no text book would use "^" to mean exponentiation! It would help for you to read responses carefully before you discount them.
    3. You reduce people's desire to help you with things like "Anyone have anything USEFUL to add?" EVERY response was helpful- unless you are expecting someone to just give you the answer.
     
  12. Jan 29, 2005 #11
    Okkkk

    Actually, I wasn't expecting the answer, I don't even want the answer, I was looking for help on the first step. Second of all, when someone tells you that your first step is wrong so they don't even bother looking at the rest, it makes you a little irritated. Lastly, I felt that my equation was correct, so excuse me if I'm not so hip on how to make things look exactly as they are in my book on this forum. You guys think that all we want are answers but you are SOOO wrong. We only come here for a little help, not to be insulted for our lack of knowledge about mathematics of whatever else it may be.
     
  13. Jan 29, 2005 #12

    Curious3141

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    The problem can be solved by taking natural logarithms of both sides. You should know these identities, but just in case you don't here they are :

    [tex]\log(ab) = \log a + \log b[/tex]

    [tex]\log (a^b) = b\log a[/tex]

    With those, you should be able to solve the problem.
     
  14. Jan 29, 2005 #13

    Hurkyl

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    It's rare that someone gets insulted for lack of knowledge -- the usual culprit is the attitude.

    At the very least, you've learned something about notation, the exact spot where you made your mistake, and even a suggested first step. :tongue2:

    (I would also recommend trying to work through the problem properly when you use division by 3 as your first step)
    (And I would recommend doing some review exercises involving exponents and their properties, preferably some of the more complicated ones)

    Have you since tried the problem again?
     
  15. Jan 30, 2005 #14
    sorry, i didn't mean to insult anyone..... ...... ...... I apologize for my words
    :frown:

    sometime it is a little bit irritated (i borrowed your word) when I spend my time here try to help people, and after I have everything typed and send out.... then the original poster said he is not asking what he said.... and want me to help him with the new question again.....

    I am very sorry hurting your feeling..... just let you know.... most of the ppls in this forums is very nice (not like me).... they volunteer to help you guys.... not getting pay or anything.... :!!)

    i promise i will try my best to be nice here.... :wink:
     
  16. Jan 31, 2005 #15
    I apologize also

    I don't want anyone to think that I am a little ungrateful college kid wo just wants the answers. I tell everyone how cool you guys are on here to help us out when we need it the most. Honestly, your response actually hurt my feelings, and made me feel really stupid, so I just lashed back on you. I hope you all know that we do appreciate your efforts. I will do a better job of making sure my equations are entered correctly next time I have questions. Can't we all just get along :)
     
  17. Jan 31, 2005 #16

    Curious3141

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    It's easy to miscontrue intent and feelings when we're online with no face to face. :smile:

    There was only one time I really lost it against a person asking for help - that was because the person multiple posted a rather complex question needing an involved answer on multiple forum sections (5 in total, the grade school, college, gen physics, gen math, etc. etc.) at the same time. By the time I had painstakingly finished typing out my solution on one forum, I found that someone else had similarly painstakingly typed out the same answer in the other section. When I confronted the OP about this sort of inconsiderate cross-posting, all I got was attitude and more attitude. And what irked me was that even a noob should have enough common sense not to irritate people with selfish crossposting like that.

    As long as you don't do something like that, I think it's a safe bet that everyone here will bend over backwards to help you. :wink:

    So, have you solved it yet ?
     
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