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What's the value of log 1 to the base 1 ? is it defined or not ?

  1. Oct 27, 2012 #1
    let log[itex]_{x}[/itex]y ,as a function of two variable, be defined from R[itex]^{+}_{2}[/itex] to R then is it continuous at (1,1) ? if so what's the image ? here the domain is D={ (x,y)/ x ε R[itex]^{+}[/itex] , y[itex]\in[/itex]R[itex]^{+}[/itex] }. like detailed discussion :).
  2. jcsd
  3. Oct 27, 2012 #2


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    Hey vrmuth.

    log_a(b) = ln(a)/ln(b) by definition but if b = 1 then ln(b) = 0 and thus the logarithm is un-defined.

    Also considering 1^anything = 1, it is no surprise that this is the case.
  4. Oct 30, 2012 #3
    oh! yes, so log is not at all defined for base 1 correct ? thanks! , is that the only discontinuity of that function ? , is it removable ? and consider log_x(1) as a function of x definded on (0,inf) then it is not continuous at x=1 , my question is can it be redefined as = 0 ? and becomes continuous ?
  5. Oct 30, 2012 #4


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    log_x(1) = log(1)/log(x) (assuming log(x) is properly defined) = 0 for all valid x and yes log_1(x) is not defined at all.

    Recall the properties that logarithms must have: you need log(xy) = log(x) + log(y) for some particular base.

    Recall that an inverse function only exists if the derivative of the original function is 0 (this is known as the inverse function theorem and it has a multi-dimensional definition).

    Now the inverse of 1^(x) is log_1(x) but the derivative of a^x is ln(a)*a^x and in this case a = 1 which means d/dx 1^x = 0 for all values of x so this means log_1(x) does not exist at all.
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