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Homework Help: Logs no caculator

  1. Jun 4, 2010 #1
    1. The problem statement, all variables and given/known data
    log_2(x)+log_4(5)=-1 just found out we have to do logs with no calculators and its been about 2 years since i last had anything to do with them. anyway i need some help to solve this.


    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jun 4, 2010 #2

    rock.freak667

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    You will need to find these. Mainly look into the product law for logs and the change of base rule.

    Also logax=logay ⇒ x=y
     
  4. Jun 4, 2010 #3
    ok ill have a look into that and get back to you if i stil cant do it... thanks
     
  5. Jun 4, 2010 #4

    Mentallic

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    The rule you need for this question are the change of base rules,

    [tex]log_ab=\frac{log_cb}{log_ca}[/tex] where c>0 and any base you choose it to be.

    Now change your log45 into a log base 2, and from there you can use your other rules of logarithms to turn the equation into [tex]log_2a=-1[/tex], so [tex]2^{-1}=a[/tex]
     
  6. Jun 4, 2010 #5
    so it will go to log5/log4 base 2 obviously and then what do i do??
     
  7. Jun 5, 2010 #6

    Mentallic

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    Yep, can that be simplified at all? What is log24?
     
  8. Jun 5, 2010 #7
    =log_2 2^2 = 2??
     
  9. Jun 5, 2010 #8

    Mentallic

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    Correct, so now we have [tex]log_2x+\frac{log_25}{2}=-1[/tex]

    Now use the rule that [tex]a.log_bc=log_b(c^a)[/tex]
     
  10. Jun 5, 2010 #9
    ok im stumped here i cant get it into a form where that can be used. log_2x=-1-log_2 5/2???
     
  11. Jun 5, 2010 #10

    Mentallic

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    No no, use it on that one term [tex]\frac{log_25}{2}[/tex]. Think about what the 'a' in [tex]alog_bc[/tex] is here.
     
  12. Jun 5, 2010 #11
    i think the a is 1/2. from that =log_2 (5^.5) still not sure how to proceed from here though????
     
  13. Jun 5, 2010 #12

    Mentallic

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    Good, now look at what you have:

    [tex]log_2x+log_2(\sqrt{5})=-1[/tex]

    The left part be simplified into one log :smile:
     
  14. Jun 5, 2010 #13
    ok i got it from here thanks for that.
     
  15. Jun 5, 2010 #14

    Mentallic

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    You're welcome.
     
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