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Can you distribute logarithms?

  1. Aug 11, 2013 #1
    Does anyone know if the following is true:

    logb (x + y) = logb x + logb y

    Thanks. This isn't homework, but I am just wondering if the following is true. I already know the logarithm product, quotient, and power rules!
     
    Last edited: Aug 11, 2013
  2. jcsd
  3. Aug 11, 2013 #2

    micromass

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    Why don't you try to find a counterexample?
     
  4. Aug 11, 2013 #3
    Ok, good idea. I don't know how to use a scientific calculator yet to figure out logs, but I know they are online. I will plug in some real numbers to see if the equation s true or not.
     
  5. Aug 11, 2013 #4
    Ok, I just did, using log 2 and log 3, and then again with log 8 and log 10. It worked! you can distribute logs!

    Thanks, and sorry for posting so many questions in one day.
     
  6. Aug 11, 2013 #5

    micromass

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    It shouldn't work :confused:

    Can you post what you did?
     
  7. Aug 11, 2013 #6
  8. Aug 11, 2013 #7
    log(2)=log(1+1)=log(1)+log(1)

    Is this true?
     
  9. Aug 11, 2013 #8

    Nugatory

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    What's the log of 1?
    What's the log of 2?
     
  10. Aug 11, 2013 #9

    micromass

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    I think he knows lol. He just asked it to the OP :tongue:
     
  11. Aug 11, 2013 #10

    Nugatory

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    Oops - right you are. Sorry about that.
     
  12. Aug 11, 2013 #11

    symbolipoint

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    Make and test an example. Imagine your base is 10, and x=100 and y=10,000,000.

    log10(100+10000000)=log10(100)+log10(10000000)
    Does this make sense? Does this not make sense?
     
  13. Aug 12, 2013 #12

    mathman

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    Log(xy) = log(x) + log(y). Unless xy = x+y, your equation is wrong.
     
  14. Aug 12, 2013 #13
    Hi, sorry for not posting sooner, but I was busy all day and I don't really have Internet access until I get to the library.

    Let me also say that I am an idiot!

    Yesterday, I said that you could distribute logs, so that:

    logb (x + y) = logb (x) + logb (y)

    But I made a mistake. Instead of adding, for example, 2 and 3, and taking the log of 5 and comparing that with the sum of log 2 and log 3, I multiplied 2 and 3! Thus I got:

    logb x + logb y = logb (xy)

    which is, of course, the product rule.
     
  15. Aug 12, 2013 #14

    symbolipoint

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    I did not say that the equation was correct. I only presented it, and then asked two questions. I know already that the equation is wrong. mileena already found understanding that was sought.
     
  16. Aug 12, 2013 #15
    @symbolipoint: I think mathman may have been responding to the original post when he was commenting.

    @Nugatory: log 1 = 0 and log 2 ≈ 0.6931...

    -Junaid :tongue:
     
  17. Aug 13, 2013 #16

    symbolipoint

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    mathman, that is possible. I inferred that you may have responded to my post because yours came directly after it, and other interrelations of responses of posts were not clear.

    I tried to give an example to be checked and asked if the example made sense.
     
  18. Aug 13, 2013 #17

    mathman

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    I was responding to the original post.
     
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