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Simple Logarithms

  1. Aug 22, 2007 #1
    1. The problem statement, all variables and given/known data
    #8: Simplify: log91/27

    2. Relevant equations

    y=logb(x)
    x=b^y

    3. The attempt at a solution

    I have personally not tried logarithms before, but I think the problem is really easy and im probably just over-thinking it?

    I think that, using the formula's below, 9^a=1/27
    So, a=1/243, so my simplified answer should be 9^1/243=1/27..
    Can I simplify fractions then, so that 9^1/9=1?
    I have personally no exact knowledge on what im doing, but can someone give me some hints or basic concepts on logarithms? Thanks!

    **Edit:
    Ok, I might be wrong above, I found some more laws, and since loga (x/y)= logax - loga y,
    I get:
    log 9 (1) - log 9 (27)

    since loga 1 = 0
    I know have:
    - log 9 (27)
    Err.. now what?
    I have no idea how to plot it into my calculator...
    But from my formulas (first two) I get that 9^2/3=27.. but does simplifying mean turning into exponential form? :S
     
    Last edited: Aug 22, 2007
  2. jcsd
  3. Aug 22, 2007 #2
    27 = 3^3 = 9^(3/2) ==> log_9(27) == 3/2
     
  4. Aug 22, 2007 #3
    Huh? I'm pretty sure its 2/3... maybe im wrong but could you further explain yourself?
     
  5. Aug 22, 2007 #4
    log(base9) (1/27) = x

    9^x = (1/27)

    x = log(1/27)/log(9)
     
  6. Aug 22, 2007 #5
    So log (1/27) / log (9)

    log (1/27) = - log 27 / log 9
    Can I simplify to - log 3 / log (1)
    So - log 3 / ..0?

    Hmm, so I can't simplify or my answer is undefined...

    Is my answer -log 27 / log 9..
    Im really lost here.

    How about this:
    Since log (base a) (a) = 1, does log (base a) (3a) = 3? If so, log (base 9) (27) = 3?
     
  7. Aug 22, 2007 #6
    log(base9) (1/27) = x

    9^x = (1/27) in exponential form ... since it said to simplify, i solved sorry

    type in your calculator log(1/27)/log(9) and plug into x, 9^(-1.5) = 1/27
     
  8. Aug 22, 2007 #7
    No problem. So, 9^x= 1/27 is simplified?..

    *bangs head against wall* ( I knew it wasn't that complicated, :P)
     
  9. Aug 22, 2007 #8
    so the general rule, and with the relevant equations you gave should have been enough information for you to solve it

    anyways, the way i learned to convert/simplify/solve logs is by how you read logs

    log(base) (y) = x

    the base must be raised to some power x, to get y

    logs are just inverses, one-to-one functions
     
  10. Aug 22, 2007 #9
    Thanks alot for your help, ill go on to the next log problem :) hehe, I guess it wasn't so complicated, I just really didn't know what they ment by "simplify" but now I will assume that simplifying a log solely involves getting rid of the 'log'. Again, thanks.
     
  11. Aug 22, 2007 #10
    i just reviewed logs Monday so it's fresh on my mind!

    some can be a real b tho :) have fun
     
  12. Aug 22, 2007 #11
    Actually simply means simplify until you cannot simplify anymore. If you can reach a solution that is how far you need to go, like in this case.

    Here's how I did it (since you already have the answer :-/):

    [tex]log_9(\frac{1}{27}) = \frac{log_{3}\frac{1}{27}}{log_{3}9}[/tex]

    [tex]\frac{log_{3}\frac{1}{27}}{log_{3}9} = \frac{log_{3}3^{-3}}{log_{3}3^2}[/tex]

    [tex]\frac{log_{3}3^{-3}}{log_{3}3^2} = \frac{-3}{2}[/tex]
     
  13. Aug 22, 2007 #12

    learningphysics

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    [tex]27 = 9 * 3 = 9 * 9^{1/2} = 9^{1 + 1/2} = 9^{3/2}[/tex]

    [tex]1/27 = 27^{-1} = (9^{3/2})^{-1} = 9^{-3/2}[/tex]
     
    Last edited: Aug 22, 2007
  14. Aug 22, 2007 #13
    Wow... Thanks alot for that! That explains alot :P
     
  15. Aug 22, 2007 #14
    Umm, may I ask how you get from [tex]log_9(\frac{1}{27}) = \frac{log_{3}\frac{1}{27}}{log_{3}9}[/tex]

    Is there a formula?
    Like...: log(base a) (b) = log (base (square root of a)) (b) / log (base (square root of a)


    You kind of make sense, but I want to know how to implement it in any problem..
    I hope you don't mind explaining after you solved it...
     
  16. Aug 22, 2007 #15
    Change of base is what it is, most people don't realize you can change to any base and not just base 10.

    [tex]log_{a}b = \frac {log_{n}b}{log_{n}a}[/tex] where n is an integer and is the new base.
     
    Last edited: Aug 22, 2007
  17. Aug 22, 2007 #16
    Umm, but leaving n as an integer doesnt that give room for n=any integer while a and b dont change? Im going to analyze this and ill get back to you.
     
  18. Aug 22, 2007 #17
    change-of-base formula is ... log(b)M=log(a)M/log(a)b where log(a) = log(base10)

    let's start off with the Proof

    Proof, let x = log(b)M ... in exponential form b^x = M

    1) b^x = M

    2) take the log(a) of both sides ... log(a)b^x = log(a)M

    3) solve for x ... x = log(a)M/log(a)b

    4) if x = log(b)M ... (step 3) log(b)M = log(a)M/log(a)b

    ... log(a) can be anything, it doesn't have to be log(base10)
     
    Last edited: Aug 22, 2007
  19. Aug 22, 2007 #18
    so... if my answer is log(3) 3^-3 / log (3) 3^2 = -3/2...
    Can it also be log (5) 3^-3 / log (5) 3^2 = -3/2.. ?

    Or does the new base have to be related to the original base? This is what has me confused.. How do you define the new base, and what effect does this have on the other terms of the equation?

    This would be because even though even though the numbers may be larger/smaller, the ratio is the same? I hope im right this time, I think im getting this logs thing :)
     
    Last edited: Aug 22, 2007
  20. Aug 22, 2007 #19
    i think so ... lol, but i'm no expert.
     
  21. Aug 22, 2007 #20
    Thanks, ill re-check with my teacher tommorow. Anyways, I got stuck again.
    1. The problem statement, all variables and given/known data
    Convert [tex]\frac{8}{\log_{5}9}[/tex] to an expression of the form [tex]a\log_{3}b[/tex]
    Ok that's the problem.

    2. Relevant equations
    #1: [tex]y=\log_{a}(x) [/tex]
    #2: [tex] x=a^y [/tex]
    3. The attempt at a solution
    **Im working on it as you read :) bare with me and I'll give you my attempt hehe
     
    Last edited: Aug 22, 2007
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