Yahoo Answers is shutting down on May 4th, 2021 (Eastern Time) and beginning April 20th, 2021 (Eastern Time) the Yahoo Answers website will be in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.

why isn't the logarithm of a negative number defined? and why can't the base be negative?

-2^3 = -8

so log -8 = 3 what's wrong with that?

-2

13 Answers

Relevance
  • RickB
    Lv 7
    1 decade ago
    Favorite Answer

    It's because expressions like "-2^3 = -8" only work for powers that are integers. (Well, only for powers that are rational numbers with odd denominators. But I digress.)

    Here's what I mean:

    You could use the above equation to assert that "log (base -2) of -8 equals 3", and that works fine, because it's easy to see that -2^3 = (-2)(-2)(-2) = -8. And you can assert that "log(base -2) of 16 equals 4", because it's easy to see that -2^4 = 16.

    But what if somebody asked, "how much is log(base -2) of -16?" To what power can you raise -2 to get "-16"? Likewise, to what power can you raise -2 to get "+8"?

    If you graph "y = (+2)^x" you get a nice continuous curve. Take its inverse, and you get a nice continuous logarithm curve.

    But try graphing "y = (-2)^x".

    * If x is an odd integer, y is negative.

    * If x is an even integer, y is positive.

    * if x is a fraction with an odd denominator, y will be positive or negative, depending on the numerator.

    * If x is a fraction with even denominator, y is undefined (because t requires you to take an even root of -2)

    * If x is NOT a fraction or an integer (like (-2)^π), who's to say whether y is positive or negative? or defined?

    So, the graph of "y = (-2)^x" is not a nice curve, but a discontinuous jumble of dots that jumps from positive to negative and which is not defined for most numbers. Its inverse would not make a very good logarithm function.

  • 1 decade ago

    http://en.wikipedia.org/wiki/Common_logarithm

    Thus note: For basic mathematics, logarithm is interpreted as having base 10. While with more complex mathematics, logarithm is now with the natural base, e. See the 'notations' section on the above wiki.

    The common logarithm does not define those with negative values. Thus there is no logarithm of -8 to base 10.

    Logarithms can also have different bases (other than 10 or e) but again, they can only be positive bases. The problem with negative ones is that they may yield several values and they do not follow many of the usual laws. Thus there is no logarithm of -8 to base -2.

    Meanwhile for the natural base (we write that as ln x), the logarithm of any nonzero number is defined! Primitive value only.

    (That means, they had defined the logarithm of -2 to base e. In fact the primitive answer is ln (-2)= (ln2)+πi. i is the imaginary number, i²=–1.)

    Unfortunately, there is a breakdown of rules in complex logarithm.

    See http://en.wikipedia.org/wiki/Complex_numbers_expon...

    This means ln -8 is not equal to 3*{ln -2}.

  • ?
    Lv 4
    4 years ago

    Negative Number Definition

  • Helmut
    Lv 7
    1 decade ago

    Since (-2)(-2)(-2) = -8, you can say that log{base-2}(-8) = 3, but what will you do then about positive numbers? You run into the same difficulty.

    Another, different, problem arises with the exponential definition of logs of negatives numbers. While e^iπ = -1, so is e^-iπ, and, indeed, e^iπ(2n+1), so you end up with an ambiguous multivalued function

    ln(-1) = iπ(2n + 1) has an infinite number of values.

  • How do you think about the answers? You can sign in to vote the answer.
  • Anonymous
    1 decade ago

    actually, the logarithm of negative numbers exist on the complex number plane.

    The Natural Logarithm of negative one is equal to the constant pi multiplied by the imaginary number i

    ln (-1) = πi

    A famous formula states that e^(i π) + 1 = 0

  • Anonymous
    1 decade ago

    That is a good question. Here is my guess.

    logs were invented to be the inverse function of exponental functions, and by definition the base is greater than zero.

    If you expanded the definition to include bases with negative values then you would not have a continous exponential function, in fact many of the values, say like N = (-2)^2.9 would not exist, so we would 'know' that log N = 2.9 without knowing what N is.

    If anyone reading this with a deeper understanding of math knows what -2^2.9 corresponds to, please add to this discussion.

  • 1 decade ago

    The equality you stated is true, but...

    When someone says "logarithm" they usually mean either base e or base 10. These particular logarithms yield no result for negative inputs. If you wanted to define a logarithm base -2, then for some negative inputs, it would have a meaningful result. If no other information is given, the default assumption is usually logarithm base e.

    It's more correct to say "the logarithm, base e, of a negative number is not defined".

  • 1 decade ago

    Because you are cheating. What you really are doing is -1 * (2^3) = -8

    We can't define the log of a negative number because, say log x = a, because as x -> 0 from the positive side, the value of a becomesincreasingly negative without bound.

  • Anonymous
    1 decade ago

    When the logarithms of negative numbers are defined, it would give rise to lots of complications.

  • 1 decade ago

    Hello,

    The log is really the power to which you raise a number to get another i.e.

    log 100 = 2 we say that 2 is the power to which you raise 10 to get 100. So log -100 = ? asks us what is the power to which we raise 10 to get -100. This cannot be done.

    Hope This Helps!!

Still have questions? Get your answers by asking now.