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Why can't logs have a negative base?

  1. Feb 21, 2016 #1
    I understand that taking logs of a negative number isn't possible because no number to any power produces a negative number. But why not a negative base?
    Say, log-10(100) = 2. Rewritten, -102 = 100, which is accurate.
    You could suggest that you may as well just ignore the negative because -x2 = x2, but it's still weird that this shows up as a syntax error on a calculator.
     
  2. jcsd
  3. Feb 21, 2016 #2

    mfb

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    Try to calculate log-10(50)=x.
    Or, alternatively, solve log-10(x)=1.7. What is x=(-10)1.7?

    For integers (with the right sign) as logarithm values, it would work, but only for those.
     
  4. Feb 21, 2016 #3

    Mark44

    Staff: Mentor

    No, it isn't. ##-10^2 = - 100## because the exponent has a higher precedence than the negation sign. What you probably meant was ##(-10)^2##, which is 100.
    For the reason given above, ##-x^2 \ne x^2##, unless x happens to be 0. If you want to square a negative number on a calculator, put parentheses around the number, with the exponent outside the parentheses.
     
  5. Feb 21, 2016 #4

    Mark44

    Staff: Mentor

    Moved from Homework sections, as this is more of a general question than a homework question. @Rumplestiltskin, be advised that if you post in the HW sections, youi need to use the homework template, not delete it as you apparently did.
     
  6. Feb 21, 2016 #5

    SammyS

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    You should be more careful regarding order of operations.

    -102 = -100

    and

    -x2 = - (x2)

    I assume you meant to write

    (-10)2 = 100

    and

    (-x)2 = x2


    The main problem with having a negative base in a logarithm, is that there is a problem defining a real valued exponential function having a negative base.
     
  7. Feb 23, 2016 #6
    Syntax error on calculator. When typed into google, (-10)1.7 = 29.4590465 - 40.5468989 i. Woah. Still at a loss.
    Could you elaborate?
     
  8. Feb 23, 2016 #7

    SammyS

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    That's a complex number. Isn't that a problem for a real function?

    Try (-10)π .
     
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