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Math Help - Logarithmic differentiation

  1. #16
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by db5vry View Post
    It isn't, I am not sure how to do that at all. I feel increasingly out of my depth with this question
    Clearly, x^{-\sqrt{x}}\ne0 so that f'(x)=0 precisely when \frac{1}{\sqrt{x}}+\frac{\ln(x)}{2\sqrt{x}}=\frac{  2+\ln(x)}{2\sqrt{x}}=0 which is clearly true when \ln(x)+2=0\implies \ln(x)=-2\implies x=\tfrac{1}{e^2}.
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  2. #17
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    Quote Originally Posted by Drexel28 View Post
    Clearly, x^{-\sqrt{x}}\ne0 so that f'(x)=0 precisely when \frac{1}{\sqrt{x}}+\frac{\ln(x)}{2\sqrt{x}}=\frac{  2+\ln(x)}{2\sqrt{x}}=0 which is clearly true when \ln(x)+2=0\implies \ln(x)=-2\implies x=\tfrac{1}{e^2}.
    I see! I wasn't thinking of it in that way. I understand that ln (x) is equal to -2, but how did you see that x was equal to \frac{1}{e^2}?
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  3. #18
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by db5vry View Post
    I see! I wasn't thinking of it in that way. I understand that ln (x) is equal to -2, but how did you see that x was equal to \frac{1}{e^2}?
    \ln(x)=-2\implies x=e^{-2}=\frac{1}{e^2}
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  4. #19
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    Quote Originally Posted by Drexel28 View Post
    \ln(x)=-2\implies x=e^{-2}=\frac{1}{e^2}
    That's fantastic! I see what you were guiding me towards all along now. I need a break after a couple of hours now and want to thank you for helping me out
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