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What is the inverse of this logarithm equation?

  1. Jan 2, 2013 #1
    1. The problem statement, all variables and given/known data

    What is the inverse of this logarithm equation?

    y=-log5(-x)




    i tried it and i got y=-5(-x)


    hm.. you know how people say that if you want to find the inverse graph of something just switch the x and y coordinates from the table of values? Well i also tried that approach and apparently the inverse is y=log5x

    atleast thats how it looks on table of values and on the graph....i dunno why i went with the y=-5(-x)


    I dont know if i got it right or not, someone kind enough to take a look and see if i got it right or not? or maybe just give me the solution outright so i know if i did it correctly or not? :P
     
    Last edited: Jan 2, 2013
  2. jcsd
  3. Jan 2, 2013 #2

    symbolipoint

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    Your result is correct. If you start from your original equation, multiply both sides by negative 1, you can easily switch x and y, and carry couple simple steps to find y as a function of x.
     
  4. Jan 2, 2013 #3
    hm... so y=log5x is correct?
     
  5. Jan 2, 2013 #4

    symbolipoint

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    No! Your first result was correct. y=-5^(-x)
     
  6. Jan 2, 2013 #5
    ahh ok ty
     
  7. Jan 2, 2013 #6

    symbolipoint

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    Original equation: [itex]y=-log_{5}(-x)[/itex]

    [itex]-y=log_{5}(-x)[/itex]
    Now express exponential form :
    [itex]5^{-y}=-x[/itex]

    Now, switch x and y to create inverse:
    [itex]5^{-x}=-y[/itex],

    and then simply, [itex]y=-5^{-x}[/itex]
     
  8. Jan 2, 2013 #7

    Mentallic

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    To check if you're right, when you get to the equation [itex]x=-5^{-y}[/itex], just plug that value of x into your original equation [itex]y=-\log_5(-x)[/itex] and see if you get an equality.
     
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