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Switching between exponential and logarithmic form

  1. Apr 8, 2007 #1
    Hello you bunch of owls, I'm doing my homework at the moment and I'm curious, how woul I express the logarithmic equation

    f(x) = log5 (x) + 3 in it's exponential form (where 5 is the base).

    This isn't part of the homework, I'm just supposed to graph it, but I'm curious as to what the exponential form looks like.
     
  2. jcsd
  3. Apr 8, 2007 #2

    Gib Z

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    [tex]5^{f(x)} = 5^{\log_5 x +3} = 5^{\log_5 x} \cdot 5^3
    =125x[/tex]
     
  4. Apr 8, 2007 #3
    Umm, I'm sure you're right, but could you elaborate on why that's correct? Could you explain how you got that i mean.

    And is there a x= form of that, that's what I'vebeen trying to come up with :p.
     
  5. Apr 9, 2007 #4
    Can anyone explain this please? Or atleast tell me what to google to find out why this works?
     
  6. Apr 9, 2007 #5

    Gib Z

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    Well say I have something equal to each other. a=b.

    Then x^a is equal to x^b, since a=b. So in this case, f(x)=log_5 x + 3, I did
    5^(log_5 x+3) = 5^(f(x)), and I reversed the rule [tex]a^ma^n=a^{m+n}[/tex] on the 5^(log_5 x +3 ) and there we go :)

    And you trying to make it equal x?
    [tex]5^3x=5^{f(x)}[/tex]
    [tex]x=5^{f(x)-3}[/tex]
     
  7. Apr 9, 2007 #6
    Ahh, alright, thank you very much Sir :)
     
  8. Apr 9, 2007 #7

    Gib Z

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    Thats alright, but please don't call me sir. im 15 years old lol, Sir makes me feel like im 40 >.<"

    EDIT: Not that theres anything wrong with being 40 !!!:P

    EDIT 2: ..OR OLDER...damn political correctness..
     
    Last edited: Apr 9, 2007
  9. Apr 9, 2007 #8
    Could you tell me how you came up with 125x?

    P.S.:haha
     
    Last edited: Apr 9, 2007
  10. Apr 9, 2007 #9

    Gib Z

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    [tex]5^{f(x)} = 5^{\log_5 x +3} = 5^{\log_5 x} \cdot 5^3
    [/tex]. You should be able to follow that so far. Now, by definiton of the logaritim, [tex]a^{\log_a x} =x[/tex]. And 5^3 is just 125 by expanding it..
     
  11. Apr 9, 2007 #10
    Okay, thanks a bunch:)
     
  12. Apr 9, 2007 #11
    53=125

    /*extra characters*/

    EDIT: I guess Gib Z beat me to it.
     
  13. Apr 9, 2007 #12

    Gib Z

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    Lol just 16 minutes late d_leet :P
     
  14. Apr 9, 2007 #13
    Eh, it's been a long day.
     
  15. Apr 9, 2007 #14

    HallsofIvy

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    loga(x) and ax are inverse functions.
    If y= loga(x) then x= ay and vice-versa.
     
  16. Apr 9, 2007 #15
    Last edited: Apr 9, 2007
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