1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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 ?

  2. jcsd
  3. Feb 13, 2012 #2


    User Avatar
    Homework Helper


    [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


    User Avatar
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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]

    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


    User Avatar
    Homework Helper

    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]


    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


    User Avatar
    Homework Helper

    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


    User Avatar
    Homework Helper

    No worries, and sorry again about wasting your time and confusing you with all that nonsense before.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook