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Logarithmic into exponential form

  1. Apr 17, 2007 #1
    1. The problem statement, all variables and given/known data

    Write this logarithmic form into exponential form

    2. Relevant equations

    N = 0.2 Log4 V

    3. The attempt at a solution

    I tried

    V/0.2 = 4^N

    doesnt seem to work.

    I've tried doing various inverse functions but that doesnt seem to work either, giving me a headack :(
     
  2. jcsd
  3. Apr 17, 2007 #2

    cepheid

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    Hi

    I can't understand how you could have obtained the solution you wrote. That having been said, it's the same method for all of these types of questions, right? How can you get rid of the logarithm? By applying the inverse operation, which is taking the exponential. Remember also that what you do to one side of the equation, you must do to the other.
     
  4. Apr 17, 2007 #3
    log p^m can be written as m log p.
     
  5. Apr 17, 2007 #4
    lets say for example

    4 = log3 81

    that in exponential form would equal 81 = 3^4

    so that means for Y = logb X

    The exponential form is X = B^Y

    so when i am giving the equation N = 0.2 log4 V

    when chaning that to exponential form
    does that give me

    V = (0.2)(4^N) ???

    i just dont get it man...
     
  6. Apr 17, 2007 #5
    no, you either make it N/0.2=logV or N=logV^0.2.
    You need to re-read the basics in your textbook.
     
  7. Apr 17, 2007 #6
    i think i am starting to get it....

    In my book it only talks about Y = logb X

    It didnt say anything about Y = M logb X

    I think i see the picture now.

    Thanks for the replys.
     
  8. Apr 17, 2007 #7
    Is T = n + log2 N
    the same as T = Nlog2 N???

    How would you write T = n + log2 N
    as a exponential form

    would it be 2^t-2 = N

    or would it be 2^t = N^N

    or would it be 2^T/n = N

    :(???
     
  9. Apr 17, 2007 #8

    cristo

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    no

    Firstly, you isolate the logarithm; [itex]T-n=log_2 N[/itex]
    Now, raise each side to the power 2; [itex]2^{T-n}=N[/itex]
     
  10. Apr 17, 2007 #9

    Okay i understand this

    but this leads to a more complicated problem

    T - N = log2 N

    The exponential form therefore equals = 2^x-y = y

    NOW, if the question ask you to sketch a graph for this function, how would you do it??, it would be impossible because in order to get the value of "y" you need to know what the exponent x-y is. But you wont know what x-y is because you dont know the value of y.

    so lets say for a x value of 1 you would go 2^1-y = y ..... then i am stuck....

    so whats there to do....? :(


    p.s thxs for the replys by the way, i feel this is making progress ( i have clicked on the top adverstisment 10 times already to help support this forum )
     
  11. Apr 17, 2007 #10
    You could write 2^x-y=y as 2^x/2^y=y, or 2^x=y2^y
     
  12. Apr 17, 2007 #11

    cepheid

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    In case that's hard to read,

    [tex] 2^{x-y} = 2^x2^{-y} = y [/tex]

    [tex] \Rightarrow 2^x = (2^y)y [/tex]
     
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