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How can I solve:(x^2)(e^x) - e^x = 0and2e^(2+x)=6For

  1. Aug 15, 2007 #1
    How can I solve:

    (x^2)(e^x) - e^x = 0



    For the first one, tell me if this is right:

    e^x(x^2-1)=0 ->
    e^x = 0 and x^2 - 1 = 0

    so x = 1 and 0? but 0 doesn't work when I plug it back in. so is 1 the solution for x?
  2. jcsd
  3. Aug 15, 2007 #2
    You probably meant the right thing (the follow-up text indicates so). Nevertheless I'd like to point out that the correct statement would be "... OR ...", since in mathematics saying "and" implies that both conditions have to be met (simultaneuously), whereas you (hopefully) meant that x must meet at least one of the conditions.

    I assume you get x=1 from x²-1=0. Are you sure that is the only value x which makes that condition true?
    I can only guess where you got x=0 from. It doesn't really matter since x=0 does not satisfy any of the two conditions: [tex] e^0 = 1 \neq 0, \ 0^2 -1 = -1 \neq 0 [/tex].
    Last edited: Aug 15, 2007
  4. Aug 15, 2007 #3

    Well, x can then equal 1 and -1, if x²-1=0.

    Am I at least doing this part correctly, deriving the right answer? Was I correct to factor out the e^x? Because then I'm left with:

    e^x(x²-1)=0 and then can't I get:

    e^x = 0 or x²-1 = 0?

    Then how can I figure out e^x = 0? Is there a way?

    And for the second one, this is what I tried doing:

    2e^(2+x)=12 <- this was the proper question, typo in the first post

    e^(2+x) = 6
    (2+x) ln e = ln 6
    (2+x) = ln 6
    x = ln6 - 2

    So does x = ln6 - 2????? I hope I did that right....
  5. Aug 15, 2007 #4

    [tex]\ln_e{e^x}_ = \ln{0}[/tex]
    [tex] x = \ln{0}[/tex] -- Which is undefined so it cannot be used as an answer.

  6. Aug 15, 2007 #5
    REALLY??? OK good :)
  7. Aug 15, 2007 #6


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    It is true that x^2 - 1 = 0 if and only if (x = 1 or x = -1).

    x, of course, cannot be both 1 and -1.

    If you know that x^2 - 1 = 0, you are still not guaranteed that x = 1 is a solution to whatever problelm you are trying to solve. Similarly, you are not guaranteed that x = -1 is a solution. You are simply guaranteed that one of them is a solution. If you aren't even sure that x^2 - 1 = 0 is the answer -- e.g. you are only considering it as a possible constraint -- then it still might be the case that neither x = 1 nor x = -1 are solutions to your problem.
  8. Aug 15, 2007 #7

    How come it can't be both?

    x^2 e^x - e^x = 0

    If I plug in either 1 or -1, it still gives me 0 as a solution, does it not?

    1^2 e^1 - e^1 = 0
    e^1 - e^1 = 0 <- right?

    and also the same with -1:

    -1^2 e^-1 - e^-1 = 0
    e^-1 - e^-1 = 0 <- right?
  9. Aug 15, 2007 #8
    He's saying it's a constraint because when dealing with "e and natural logs" for some numbers they are undefined. It's always a good idea to plug in your answers into the original equation to make sure they are truly answers. In this example the answer is 1 or -1 since both work in the original equation (x^2)(e^x) - e^x = 0. Does that make sense?
  10. Aug 15, 2007 #9
    Can I say for an answer:

    x equals 1 AND -1

  11. Aug 15, 2007 #10


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    No. You should say x=1 OR x =-1.... x cannot equal two numbers at the same time... it is true that x=1 satisfies the original equation AND x=-1 satisfies the original equation... but when you say x=1 and x=-1, you're saying that x=1 and at the same time x=-1...

    Suppose I tell you that I'm thinking of a number... call it x. you ask me is x=1? I say yes. Then you ask is x=-1? If I said yes, then that means that there's a number that is 1 and -1 at the same time... but that's obviously not true since 1 and -1 are different numbers.

    EDIT: think of it like this x is an actual number that I've thought up... and I tell you that this number x has the property that:

    (x^2)(e^x) - e^x = 0

    but I don't tell you any more. so you solve the equation and arrive at the conclusion that x=1 or x=-1... but you can't narrow it down further.
    Last edited: Aug 15, 2007
  12. Aug 15, 2007 #11
    No x^2-1 = 0 means that x CAN be 1 or -1, but you cannot narrow down the answer anymore then that. So you say the x=-1 or x=1 since you cannot be sure.
  13. Aug 15, 2007 #12


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    If x = 1 AND x = -1, then it would follow that 1 = -1.

    Plain english is slightly ambiguous here -- the phrase

    x = 1 and x = -1 is a solution to x^2 - 1 = 0

    has two interpretations. The following is correct:

    x = 1 is a solution to x^2 - 1 = 0, and x = -1 is a solution to x^2 - 1 = 0.

    The following is incorrect:

    (x = 1 and x = -1) is a solution to x^2 - 1 = 0.

    Well, if the author is careful about his grammar, you can tell the difference between the two; the former would be written "are solutions", not "is a solution".

    Technically speaking, (x = 1 and -1) isn't even a grammatically correct phrase!

    I admit that I'm being somewhat pedantic here -- but I think that's important. Experts can get away with being sloppy, but sloppiness tends to confuse students.
    Last edited: Aug 15, 2007
  14. Aug 15, 2007 #13
    I don't know why but it seems like physics people are always striving to be more precise then strictly math people... :rofl:
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