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ameni

  • 3 years ago

f'(x)=(2 ln(x))/x How do I find the x value? I put the equation equal to zero, but I do not know any log rule to solve it for x value. I need help please.

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  1. Walleye
    • 3 years ago
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    Do you want to solve the equation for x?

  2. ameni
    • 3 years ago
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    Yes

  3. across
    • 3 years ago
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    That's not possible, unless you're setting f'(x)=0.

  4. across
    • 3 years ago
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    Are you trying to find maxima/minima?

  5. across
    • 3 years ago
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    \[\frac{2\ln(x)}{x}=0\]doesn't seem too bad.

  6. imranmeah91
    • 3 years ago
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    if you are looking for critical value , look for where function diverge too

  7. ameni
    • 3 years ago
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    How do I proceed after this to find the value for x?

  8. ameni
    • 3 years ago
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    Yes I am looking for critical values

  9. imranmeah91
    • 3 years ago
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    what happened if you plug in 0 for x

  10. ameni
    • 3 years ago
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    you can't take the ln(0). There has to be some other way related to log rules to solve this question, and I can't figure it out.

  11. imranmeah91
    • 3 years ago
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    it diverges, which mean it is a criticle point. a critical point of a function of a real variable is any value in the domain where either the function is not differentiable or its derivative is 0

  12. across
    • 3 years ago
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    \[\frac{2\ln(x)}{x}=0,\]\[2\ln(x)=0,\]\[\ln(x)=0,\]\[x=1.\]

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