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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


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    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]
    [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]
    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


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    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]
    [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


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    [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.
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