1. Not finding help here? Sign up for a free 30min 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!

Finding a constant

  1. Dec 2, 2005 #1
    Is it possible to find a constant value for b in the following equality for any value of x?

    [tex] b^{x}(\ln b) = a^x [/tex]
     
  2. jcsd
  3. Dec 2, 2005 #2
    what is a? Is it given? If so, in what field is it an element.
     
  4. Dec 2, 2005 #3
    a can be any value, so the real question is to find b in relationship to a, independantly of x.
     
    Last edited: Dec 2, 2005
  5. Dec 2, 2005 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Since you didn't answer the question "in what field is it an element?" I assume this is in the real numbers.

    [tex] b^{x}(\ln b) = a^x [/tex]
    Taking ln of both sides:
    [tex]ln(b^x ln b)= ln b^x + ln b= xln b+ ln b= x ln a[/tex]
    so
    [tex]x(ln a- ln b)= x ln\frac{a}{b}= ln b[/tex]
    That will be "independent of x" if and only if
    [tex]ln \frac{a}{b}= 0[/tex]
    or
    a= b= 1.
     
  6. Dec 2, 2005 #5
    I think you made a mistake

    You're development:

    [tex]\ln(b^x \ln b)= \ln b^x + \ln b[/tex]

    When it should be

    [tex]\ln(b^x \ln b)= \ln b^x + \ln(\ln b)[/tex]
     
    Last edited: Dec 2, 2005
  7. Dec 2, 2005 #6

    CRGreathouse

    User Avatar
    Science Advisor
    Homework Helper

    Suppose [tex]a,b\in\mathbb{R}^+[/tex]. If [tex]b^x\ln b=a^x[/tex] then we have
    [tex]x\ln b+\ln\ln b=x\ln a[/tex]
    or
    [tex]\ln\ln b=x(\ln a-\ln b)[/tex].

    If [tex]a\neq b[/tex] then this will not hold for all [tex]x\in\mathbb{R}[/tex]. Then we have a=b and [tex]\ln\ln b=0[/tex]. This means the unique solution is [tex]a=b=e[/tex].
     
  8. Dec 2, 2005 #7
    Does that mean that the function a^x has no integral function if a is not equal to e?
     
  9. Dec 3, 2005 #8

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    ???
    [tex]a^x= e^{ln a^x}= e^{x ln a}[/tex]
    so the anti-derivative (I guess that's what you mean by "integral function") is
    [tex]\frac{e^{x ln a}}{ln a}= \frac{a^x}{ln a}[/itex].

    Yes, werg22. Thanks for the correction.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?