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Change of base formula, is this what hes talking about? logs!

  1. Nov 13, 2006 #1
    Hello everyone.

    This is for a descrete math class, and he said you must know the change of base formula, so if you have everything in log base 3, you can figure out what it is in log base 2 as an example.

    But i'm looking up the examples on the interenet and they all seem to convert whatever log they have to log base 10 and evaluate from there. Is this also what he would be talking about? No calculators are allowed to evaluate the logs.

    Here is the change of base formula:
    [​IMG]

    and here is the example i'm looking at:
    [​IMG]


    Okay i see that x = 4, and z = 8, but what is the base y? is the base y the new base you are trying to convert the orginal base into? FOr this example, you are given a base 4 log, and your trying to convert it to a log of base 2 and evaluate the answer i'm assuming right?

    how did they end up with 3/2 though?

    EDIT: I figured out how they got 3/2, i forgot about the other form of logs, 2^3 = 8, and 2^2 = 4.

    Another example:
    Use the change of base formula to evaluate
    [​IMG]


    Choices:

    A. 4.56

    B. 4.86

    C. 0.21

    D. 5.16

    Correct Answer: B

    Solution:

    Step 1:[​IMG]

    Also is it up to you what base you convert it too? I see they used base 10 but there was no base 10 in the orginal problem.
    how did they get: 2.43 from log_10(29)/log_10(4) ?


    I know this is algebra stuff but its been awhile :blushing:

    Thanks~
     
    Last edited: Nov 13, 2006
  2. jcsd
  3. Nov 13, 2006 #2

    JasonRox

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    They choose 10 because most calculators have a log base 10 button, but not a log base 2 button (I'm none have this).

    Another choice is base e, which is what I would use personally. All scientific calculators have a button for log base e, and it is usually understood as ln.

    That's the real reason.

    Sure, you can use base 5, but now what? You still can't solve it.

    Note: I saw that you can't use calculators, then I'm assuming you are allowed tables or something to figure it out.
     
  4. Nov 13, 2006 #3

    JasonRox

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    Oh, since it is multiple choice, you can just check each answer by brute force. Not that hard.
     
  5. Nov 13, 2006 #4
    we won't be allowed to use tables or calculators so maybe i'll switch everytyhing to base 2?
     
  6. Nov 13, 2006 #5

    JasonRox

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    No, just solve by brute force. I can hardly call it force though.

    You have multiple choices, so you can simply check which is correct one by one.

    The solution is probably just to prove that it is the solution.
     
  7. Nov 14, 2006 #6
    On the exam it won't be multiple choice though, this was an example i found on the internet to help me study. His exams usually arn't multiple choice but maybe i can hope hah.
     
  8. Nov 14, 2006 #7

    JasonRox

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    Oh!

    I see it now.

    The one online isn't a good example though because it doesn't have a "nice" solution to it.

    The one your teacher gave does have a nice solution. Notice how he appropriately chose 2 as the base?

    Here, I made a "nice" solution question for you...

    Solve the following without calculator or tables:

    [tex]log_{243}27[/tex]
     
  9. Nov 14, 2006 #8
    i got log_3(27)/log_3(243) = 3/5

    look good?
     
  10. Nov 14, 2006 #9

    JasonRox

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    BRAVO!

    Now, from that exercise, you probably know how I constructed it.

    I just picked a number like 6/7 and worked backwards. Once you understand how I did it, you will have no problem solving them in the future.
     
  11. Nov 14, 2006 #10
    thanks that did help! :)
     
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