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Need help understanding logarithmic properties

  1. Feb 1, 2012 #1
    Okay, so in class I learned that:

    10^log(4x) = 4x

    But I don't understand why.

    I get that a log without a subscript is considered base 10, so:

    10^y = 4x

    Is the way to understand "log(4x)", right?

    What if the problem was a different base? Would the "10" coefficient and 'log' in the original problem disappear also? If not, what coefficient would it have to be to match the base log in order for those to disappear?
     
  2. jcsd
  3. Feb 1, 2012 #2

    Well, log(4x) is the power that you raise 10 to in order to get 4x.

    And then you do it ... you raise 10 to that power ... so that 10^log(4x) = 4x.

    In other words, log(4x) is some quantity that IF you raised 10 to that power, you'd get 4x. And then you raise 10 to that power.

    A more sophisticated way of understanding this is that taking a log (with respect to some base) and raising that base to a power, are operations that are inverse to one another.

    Keep doing problems, this will become clear with practice.

    It has to be the same base. So 3^(log_3(x)) = x

    where log_3 is the base-3 logarithm.

    Because log_3(x) is the power you'd have to raise 3 to in order to get x; and then you raise 3 to that power, so you get x.
     
  4. Feb 1, 2012 #3
    Thanks, this part is what actually helped me understand it after plugging in values:

    3^(log_3(x)) = x

    3^(log_3(9)) = 9

    3^(log_3(9) = [3^?=9]) = 9
    3^(log_3(9) = 2) = 9
    3^2 = 9
    9 = 9
     
  5. Jul 14, 2012 #4

    tiny-tim

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    hi daigo! :smile:

    you seem to have difficulty in understanding equations :redface:

    some equations are definitions, and you just have to learn them

    here's a formula that may help in this case, and is easy to visualise and remember …

    ##\log_ax = \frac{\log x}{\log a} = \frac{ln(x)}{ln(a)} = \frac{\log_bx}{\log_ba}## for any bases a and b :wink:

    (and, as you know, alogax = x by definition)
     
  6. Jul 14, 2012 #5
    If it helps, think about the log function as what it is - the inverse of exponentiation, by that I mean;
    if we define [itex]f(x)=10^x[/itex], then [itex]log(x)=f^{-1}(x)[/itex], and by the definition of an inverse [itex]f^{-1}(f(x)) = x[/itex]

    To give you a few examples take;
    [itex]f(2) = 10^2 = 1000[/itex]
    so [itex]log(1000) = f^{-1}(1000) = 2[/itex]

     
    Last edited by a moderator: Jul 14, 2012
  7. Jul 14, 2012 #6

    HallsofIvy

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    For any positive number, a, [itex]log_a(a^x)= x[/itex] and [itex]a^{log_a(x)}= x[/itex]. Those essentially follow from the fact that "[itex]log_a(x)[/itex]" and "a^x[/itex] are inverse functions.
     
  8. Jul 14, 2012 #7

    HallsofIvy

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    I'm sure this was a typo but [itex]10^2= 100[/itex], not 1000.

     
  9. Jul 14, 2012 #8
    yes haha, I was pretty tired when I typed that
     
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