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Expressing this as one log

  1. Nov 14, 2012 #1
    Everything is base 10

    log(ab)-2logb -1

    a+b/ b2 = a/b/10 = 10a/b?


    But in the back of the book the answer is 10b/a...
     
  2. jcsd
  3. Nov 14, 2012 #2

    symbolipoint

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    This is an expression: log(ab)-2logb -1
    What do you want from it?

    The rest of the stuff you wrote is unintelligible.
     
  4. Nov 14, 2012 #3

    HallsofIvy

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    He stated that in the title, not in the body: express as a single iogarithm.

    That would be log((a/10b)
     
  5. Nov 14, 2012 #4

    ehild

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    Are you sure? :tongue: One parenthesis missing, is b in the numerator or in the denominator?

    ehild
     
  6. Nov 14, 2012 #5

    HallsofIvy

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    Oh, dear, an extra parenthesis! It should be log(a/10b). The "b" is in the denominator. Prior to canceling, log(ab)- 2log(b)- 1= log(ab)- log(b^2)- log(10)= log(ab/10b2).
     
  7. Nov 14, 2012 #6

    ehild

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    A missing parenthesis: log(ab)-log(b2)-log(10)=log(ab/(10b2))=log(a/(10b)) :tongue2:

    ehild
     
  8. Nov 14, 2012 #7
    Oh.. my order of operation was wrong.. thanks.
     
  9. Nov 14, 2012 #8

    Mark44

    Staff: Mentor

    Not only that, but your first post is extremely unclear as to what you're trying to do.
    1. Connect expressions that have the same value with =.
    2. Keep track of what you're doing. In the first line above, you have two log expressions. In the second line, you show no indication that you're working with logs.
    3. Indicate exponents so that we can tell what you mean. At the very least, use ^ to indicate an exponent, as in b^2. Even better would be to use the Advanced Menu (click Go Advanced, and use the X2 button, which adds HTML tags for exponents.)

    log(ab)-2logb -1
    = log(ab) - log(b2) - log(10)
    = log(ab/b2) - log(10)
    = log(a/b) - log(10)
    = log(a/(10b))
     
  10. Nov 14, 2012 #9
    = log(a/b) - log(10)
    = log(a/(10b))

    for this part I'm kind of confused if it's log(a/b)/10 shouldn't you invert and multiply and get log(10a/b)?
     
  11. Nov 14, 2012 #10

    Mark44

    Staff: Mentor

    That's not what I wrote. What you have here is
    $$ \frac{log(a/b)}{10}$$

    What I wrote is
    $$ log(\frac{a/b}{10})$$

    I hope that you can see that these are different.

    Your confusion here seems to be with basic arithmetic, particularly how fraction division works.

    $$ \frac{a/b}{10} = \frac{a}{b} \cdot \frac{1}{10} = \frac{a}{10b}$$
     
  12. Nov 14, 2012 #11

    Mark44

    Staff: Mentor

    If, as you say, the book gives the answer as 10b/a, then either you haven't written the problem correctly or the book's answer is wrong.
     
  13. Nov 15, 2012 #12
    oh... I guess I need to do some work on basic algebra......
     
  14. Nov 15, 2012 #13

    Mark44

    Staff: Mentor

    Seems like a good idea to me.
     
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