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Y=e^xlna increases less quickly than y=e^x when a<e e. i. lna<1?

  1. Feb 13, 2012 #1
    1. The problem statement, all variables and given/known data

    I find this a bit tough to digest. Can somebody, please, elaborate?

    "Because lna<1 when a<e you can see that the graph increases less quickly than the graph of e^x.."

    2. Relevant equations
    3. The attempt at a solution

    Shouldn't e^xlna > e^x even if lna<1 ?

    Thanks.
     
  2. jcsd
  3. Feb 13, 2012 #2

    Mentallic

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    Is

    [tex]\frac{1}{2}e^x>e^x[/tex] ??

    Of course not, because whatever ex is (we would determine its value if we were given x) half of it is always less since ex>0 for all x.

    You can't extend this concept for all problems however. It wouldn't be completely true that [tex]x\ln(a)<x[/tex] for 0<a<e because if x = -2 for example, then half or any other value of ln(a), [itex]-2\ln(a)>-2[/itex]

    So keep in mind when doing these problems that you're dealing with a special case that ex>0 for all x.
     
  4. Feb 13, 2012 #3
    Is e^xlna=lna*e^x ?

    So, if a=1 and lna=0, then e^xlna<e^x, because e^xlna will equal 0 ???

    Thank You.
     
  5. Feb 13, 2012 #4

    Mentallic

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    Yes, they're equivalent. I prefer to have the log function at the end though just because people can confuse lna* ex with [itex]\ln(a*e^x) = \ln a+\ln (e^x) = \ln a + x[/itex] so if you do it that way, just be sure to add parenthesis in the right places, mainly ln(a)*ex


    Yep! And if a=e then lna=1 so ex*lna = ex in this case.
    Also remember the rules of logs that you can only take the log of a number greater than zero, so when we say exlna < ex we need to add that this is only true for 0<a<e
     
  6. Feb 13, 2012 #5
    Thank You, Mentallic. It made my day.
     
  7. Feb 13, 2012 #6

    Ray Vickson

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    Since you fail to use brackets, it is not clear what your expression e^xlna means. Is it [tex] e^{x \ln(a)} \mbox{ or } e^x \ln(a)?[/tex]

    RGV
     
    Last edited: Feb 13, 2012
  8. Feb 13, 2012 #7
    The first one. E to the power of X times natural log of A. Sorry for the ambiguity.
     
  9. Feb 13, 2012 #8

    Mentallic

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    Oh... This whole time I was going with the second one :yuck:

    [tex]e^{x\ln(a)}[/tex] is equivalent to [tex]\left(e^{\ln(a)}\right)^x[/tex] :wink:


    edit: Something I have to note because I was going on the assumption that you were talking about [tex]e^x\cdot \ln(a)[/tex]

    No.

    Sorry about getting the wrong ideas into your head. Keep in mind that everything I've been saying was intended to address the question of why [tex]\ln(a)\cdot e^x < e^x[/tex] for 0<a<e

    The answer you're looking for however is in the hint I provided above about using the indice laws.
     
    Last edited: Feb 13, 2012
  10. Feb 13, 2012 #9
    [tex]\left(e^{\ln(a)}\right)^x[/tex] = a^x here?
     
  11. Feb 14, 2012 #10
    Please, check to see if this is what I was trying to see from the get-go.

    1. e^[x*ln(a)] <e^x if a<e

    If a=2 and ln(a)=0.69, then

    e^[x*ln(a)]: x=2, y=4 (approx)

    e^x: x=2, y=8 (approx)

    Here, y=e^x grows faster than y=e^[x*ln(a)]

    2. e^[x*ln(a)] > e^x if a>e

    If a=3 and ln(a)=1.1, then

    e^[x*ln(a)]: x=4, y=81 (approx)

    e^x: x=4, y=54(approx)

    Here, y=e^[x*ln(a)] grows faster than y=e^x
     
  12. Feb 14, 2012 #11

    Mentallic

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    Yes, and since a<e, [tex]a^x<e^x[/tex] which is essentially all you need to understand.

    y=4 exactly, not approximately. If a=2 then [tex]e^{x\ln(a)}=\left(e^{\ln(a)}\right)^x=a^x=2^2=4[/tex]

    Yes, pretty much. Using a few test values can help you get the idea of what is happening, but don't use it as an absolute proof.
     
  13. Feb 14, 2012 #12
    Simple and elegant. This settles it. Thank You again, Mentallic.
     
  14. Feb 14, 2012 #13

    Mentallic

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    No worries, and sorry again about wasting your time and confusing you with all that nonsense before.
     
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