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Log Base 1

  1. Aug 7, 2008 #1
    I know that logarithms to the base 1 is undefined, due to the reason that:

    [itex]\log_1{x} = \frac{\log_a{x}}{\log_a{1}}[/itex]

    And this leads to divison by zero, which is undefined.

    There was a question in one of my textbooks that asked describe the graph that results if [itex]y = \log_1{x}[/itex]. Is such a graph even possible?

    If I switch this logarithm to exponential form I would get:

    [itex]1^y = x[/itex]

    Now, is it possible that the graph could be y = 1 and x = 1? Since [itex]1^y = x^1[/itex], [itex]y = 1[/itex] and [itex]x = 1[/itex].

    Or is it the point of intersection of these two lines? If not, what is it?

    Thanks.
     
    Last edited: Aug 7, 2008
  2. jcsd
  3. Aug 7, 2008 #2

    HallsofIvy

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    Well, not "the" graph. You have two "functions" there and so two graphs. Yes, if y= log1(x) then x= 1y= 1 for all x. The reason I put "functions" in quotes above is that this is not actually a function: strictly speaking, log1(x) is not a function. But it is a relation and its graph is the vertical straight line x= 1.

    y= 1x= 1 is a function: its graph is the horizontal straight line y= 1.
     
  4. Aug 7, 2008 #3
    Well not even going into division by zero when considering a change of base, you should recognize that 1 raised to a power won't get you x, a variable.

    1 raised to any power, whether fractional, negative, or worse, is just 1. So the graph will look like x = 1.
     
  5. Nov 20, 2008 #4
    Wouldn't the expression y=log11 (or y=log1(x);x=1 for all x) be undifned too as logn1 is always zero; So y=log11 has both zero and one as an answer?
     
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