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What power do you raise 3 to to get 9

  1. May 11, 2005 #1
    Hello again..
    Alright Im now on the part of 12th grade calculus, dealing with logs and exponents and stuff. I understand if say log39 That its basically what power do you raise 3 to to get 9. Well a question i need to answer is log162. So what power to I raise 16 to to get 2??? I used my calculator just trying to find a number that would work, and i got 0.25. How would I show work for that? all of the solutions they gave for different equations worked different like for the log39They showed 2 ways:
    3^x=9
    3^x=3^2
    x=2
    OR
    log39=log3(3^2)
    =2
    None of these solutions work for 16 since 2 is 2^1 and 16^1 is 16 not 2... Im confused.. Is 0.25 even the right answer.
    Any Help is appreciated greatly!!!
    Thanks ya!!!
     
    Last edited: May 11, 2005
  2. jcsd
  3. May 11, 2005 #2
    hint: 16=2^4
     
  4. May 11, 2005 #3

    dextercioby

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    Yes.Use this property of the logarithms wrt a change of basis.

    [tex] \log_{16}2=\frac{\ln 2}{\ln 16}=\frac{\ln 2}{4\ln 2}=\frac{1}{4}=0.25 [/tex]

    Daniel.
     
  5. May 11, 2005 #4
    Have you heard of a change of base? You calculator can only do log base 10. Here is the formula

    [tex]\log_b a = \frac {\log_c a}{\log_c b}[/tex] where c can be anything. But why not set that to 10 so your calculator can crunch it.
     
  6. May 11, 2005 #5

    chroot

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    0.25 is the correct answer, because 16^(0.25) = 2.

    The easiest way to compute logs like [itex]\log_b x[/itex] on a calculator is to compute

    [tex]\log_b x = \frac{\log x}{\log b}[/tex]

    where the logs on the right can be any base at all (10, e, whatever). In other words, to find [itex]\log_{16} 2[/itex] on your calculator, punch up (log 2 / log 16).

    - Warren
     
  7. May 11, 2005 #6
    Thanks, but what does the In stand for...
     
  8. May 11, 2005 #7

    dextercioby

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    I don't think so.Any (scientifical) calculator should have natural logarithm.It's eseential.

    Daniel.
     
  9. May 11, 2005 #8

    shmoe

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    That's probably "ln", the "natural logarithm", and it means the logarithm to the base e=2.1718..., if you haven't met it in your class yet, it's probably not far off.
     
  10. May 11, 2005 #9

    chroot

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    Dexter's post using natural logs can just as well be done with logs of any base, including 10.

    - Warren
     
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