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Increasing or decreasing - logarithms

  1. Apr 25, 2014 #1

    Rectifier

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    Hey there!
    How do I determine whether this function is growing or decreasing without using any calculator or graphing tools?
    [tex]y=ln(\frac{1}{x})[/tex]
    I know that
    [tex]y=ln(\frac{1}{x}) \ \Leftrightarrow \ e^y=\frac{1}{x} \ \ \ (1) [/tex]
    Then I tried to use the formula
    [tex]f(x)=a^x[/tex] and then determine whether a is 0<a<1 or a>1. But I dont know how I should do to write (1) in the same same manner :( .

    If a is 0<a<1 means that the function is strictly decreasing.
    If a is a>1 means that the function is strictly increasing.

    Could somone please help me?
     
  2. jcsd
  3. Apr 25, 2014 #2

    Dick

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    Your function doesn't have that form. Is this a calculus class? Can't you just take the derivative?
     
  4. Apr 25, 2014 #3

    Rectifier

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    I dont know if its calculus or not. Its called one-dimensional analysis and its a class I take where I live - in Sweden.

    There should be another way to do that than the derivative since I must be able to plot the function after that without a calculator or a graphing tool.

    I am sorry if this is the wrong subforum.
     
  5. Apr 25, 2014 #4

    Dick

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    I think the forum is ok. If you have 1/x=e^y then you have x=e^(-y)=(1/e)^y. That will let you plot x=f(y) as a decreasing function of y since (1/e)<1. What does your graph translate to when you think of y as a function of x?
     
  6. Apr 25, 2014 #5

    HallsofIvy

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    Yes, "analysis" is "Calculus". (In the United States we tend to use the word "Analysis" to mean "the theory behind Calculus" so a higher level course than "Calculus".)

    Why would not using a calculator or graphing tool mean you cannot take the derivative?

    The derivative of ln(1/x)= - ln(x) is -1/x. You can tell where that is positive or negative without a calculator, can't you?
     
  7. Apr 25, 2014 #6

    Rectifier

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    Thank you for your comment!

    You went through f(y) to determine whether the function is decreasing or increasing in terms of x being a funtion of y to then mirror the function on the x=y line and get the inverse which is the original function. But the nice thing here is that since the inverse is increasing this must mean that the original is increasing too.

    Did I get it right?
     
  8. Apr 25, 2014 #7

    Rectifier

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    Yeah I can :) but oficially I am not allowed to use derivatives since I am not at that chapter of the book :D.

    I like to have more than one way to the solution. Thank you for your comment HallsofIvy :)
     
  9. Apr 26, 2014 #8

    Ray Vickson

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    [tex] y = \ln \left( \frac{1}{x}\right) = - \ln(x)[/tex]
    Do you know what the graph of ##\ln(x)## looks like?

    BTW: in LaTeX you should write "\ln" instead of "ln", because it produces better-looking results: compare ##ln(x)##, entered as # # ln(x) # # (no spaces) with ##\ln(x)##, entered as # #\ln(x) # # (no spaces).
     
    Last edited: Apr 26, 2014
  10. Apr 26, 2014 #9

    Dick

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    Right idea, but x=(1/e)^y isn't increasing. I hope that was a typo.
     
  11. Apr 26, 2014 #10

    Rectifier

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    Hehe indeed I was going to write decreasing. Thank you for your help :)
     
  12. Apr 26, 2014 #11

    Rectifier

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    Oh, I didnt see that option! Thank you for pointing this out!


    Thank you yet again :)
     
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