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(log√125 + log √27 - √8)/ log 15 - log 2 = 3/2

  1. Nov 16, 2012 #1
    1)Without using tables, show that

    (log√125 + log √27 - √8)/ log 15 - log 2 = 3/2

    What i tried was

    (3/2 log 5 + 3/2 log 3 - 3/2 log 2)/ log 5+log 3 - log2

    then from here I don't know where to take it.


    2) Find the value of x if log x2/ log a^2 = log y^4/logy

    I tried this 2logx - 2loga = 4 log y- log y
    2logx = 3logy+ 2 loga

    then here I get stuck..
     
    Last edited by a moderator: Feb 6, 2013
  2. jcsd
  3. Nov 16, 2012 #2

    ehild

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    Re: Logs

    You miss some parentheses.


    ehild
     
  4. Nov 16, 2012 #3

    Mentallic

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    Re: Logs

    Since you want the problem to reduce to something simple (3/2) you should combine the log terms and simplify from there.


    If we have

    [tex]\log(x)=y[/tex] then what is x?
     
  5. Nov 16, 2012 #4
    Re: Logs

    10^y = x?
     
  6. Nov 16, 2012 #5

    Mentallic

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    Re: Logs

    Yes, so now do the same thing. Set the equation so that it is in the form [itex]\log(x)=y[/itex] and then make x the subject. Oh and y will be some complicated expression.

    And once you've done that, remember the rules

    [tex]a^{x+y}=a^xa^y[/tex]

    [tex]a^{\log_{a}(x)}=x[/tex]
     
  7. Nov 16, 2012 #6
    Re: Logs

    I'm kind of confused if I have this

    log x = 3logy - 2loga

    I duno how to get rid of log a to make it log (x) = y
     
  8. Nov 16, 2012 #7

    ehild

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    Re: Logs

    log y^4/logy=4log(y)/log(y)=4

    ehild
     
  9. Nov 16, 2012 #8
    Re: Logs

    oh.. so it's 2log x = 4 + 2log a

    x^2= a^8

    x=a^4?
     
  10. Nov 16, 2012 #9

    Mentallic

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    Re: Logs

    ehild noticed some mistakes which you should first address.

    [tex]\frac{\log(a)}{\log(b)}\neq \log(a)-\log(b)[/tex]

    What you're thinking of is

    [tex]\log\left(\frac{a}{b}\right)=\log(a)-\log(b)[/tex]
     
  11. Nov 16, 2012 #10
    Re: Logs

    Oh, Umm should it should be

    log x^2/ log a^2 = 4

    2log x = 8 log a

    log x = 4log a

    10^a^4 = 10^x

    x= a^4?
     
  12. Nov 16, 2012 #11
    Re: Logs

    And for the first one

    is it (3/2log5 + 3/2log 3 - 3/2log 2)/ (log 5+ log 3) - log 2

    = 1/2log log + 1/2log 3 - 1/2log2?
     
  13. Nov 16, 2012 #12

    ehild

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    Re: Logs

    I think you made some mistake when copying the problem. It should be

    (log√125 + log √27 - √8)/ (log 15 - log 2 )= 3/2.


    ehild
     
  14. Nov 16, 2012 #13

    Mentallic

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    Re: Logs

    Yes that's correct :smile:

    No, again, you want to simplify that expression into a simple answer of 3/2, so what you're aiming to do is to combine each log term, not split them up further. Use the [itex]\log(a)+\log(b)=\log(ab)[/itex] rule.

    Nope the [itex]\sqrt{8}[/itex] should be [itex]\log(\sqrt{8})[/itex]
     
  15. Nov 17, 2012 #14
    Re: Logs

    I still don't get it combine them? But aren't they too big? shouldn't I try to get them down to like small numbers and then try to cancel out?
     
  16. Nov 17, 2012 #15

    Mentallic

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    Re: Logs

    Yes, combine them! If you get them down to small numbers then you'll end up with the fraction you posted earlier that cannot be cancelled further.

    Use

    [tex]\log(a)+\log(b)-\log(c)=\log\left(\frac{ab}{c}\right)[/tex]

    for both the numerator and denominator and see if you notice any nice cancellations.
     
  17. Nov 17, 2012 #16

    SammyS

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    Re: Logs

    Looks like a typo.

    How about the first one should be:

    (log√125 + log √27 - log√8)/ (log 15 - log 2 )= 3/2

    As ehild said early on, you need to use sufficient parentheses .
     
  18. Nov 17, 2012 #17
    Re: Logs

    thank you guys.
     
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