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Why is the integral 1/x equivalent to log base e?

  1. Dec 3, 2011 #1
    Yes I know it's a really stupid question, please don't make fun of me. But when they first defined this new function, the integral of 1/x, and had no graph or knowledge of how it looked, how could they know it was a logarithm? More specifically, the logarithm with base e.
     
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  3. Dec 3, 2011 #2

    micromass

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    I'm guessing that they first invented the logarithm as the inverse of the exponential function. It is then easy to show that the derivative of the logarithm is exactly 1/x.
     
  4. Dec 3, 2011 #3

    AlephZero

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    Not according to http://en.wikipedia.org/wiki/Logarithm#History

    Napier "invented" logs to base 10, for the practical purpose of doing calculations.

    Euler "invented" the exponential function more than 100 year later, though of course integer powers of numbers were known thousands of year before either of those inventions/discoveries.
     
  5. Dec 3, 2011 #4

    HallsofIvy

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    pmquable, It is relatively easy to show that [itex]d(e^x)/dx= e^x[/itex]. Since ln(x) is defined as the inverse to that function, if y= ln(x), then [itex]x= e^y[/itex]. Then [itex]dx/dy= e^y[/itex]. But that tells us that the derivative of [itex]y= ln(x)[/itex] is the reciprocal of that: [itex]d(ln(x))/dx= dy/dx= 1/e^y= 1/x[/itex].

    That was what micromass really meant- not that the "logarithm" was first defined as the inverse of the exponential function but that the natural logarithm was defined that way.

    However, many text books do it the opposite way now: First define ln(x) by
    [tex]ln(x)= \int_1^x \frac{dt}{t}[/tex]
    Of course, from that definition, we have immediately that the derivative is 1/x.
    It may be that this is the definition you have seen and now you are asking "How do we know that function really is the inverse of [itex]e^x[/itex]?

    You can prove a number of things directly from that definition: that ln(xy)= ln(x)+ ln(y), that ln(1/x)= -ln(x), that [itex]ln(x^y)= y ln(x)[/itex] for example. One can also prove that ln(x) is an increasing function that maps the set of all positive real numbers to the set of all real numbers. It follows that it has an inverse function that maps the set of all real numbers to the set of all positive numbers. Let's call that function "E(x)".

    If y= E(x) then x= ln(y). If x is not 0, we can can divide both sides by x to get [itex]1= (1/x)ln(y)= ln(y^{1/x})[/itex]. Going back to the inverse function, that says [itex]E(1)= y^{1/x}[/itex] so that [itex]y= E(x)= (E(1))^x[/itex] power. That was for x not equal to 0. If x= 0, then E(0)= 1[/itex] specifically because ln(1)= 0. But any number to the 0 power is 1 so this is still [itex]E(1)^0[/itex]. In either case, that function really is the inverse of some number to the x power. Now we can either define e to be that number or, if we have already defined [itex]e^x[/itex], observe that both it and the E(x) have the same derivative and so they differ by, at most, a constant. But [itex]e^0= 1[/itex] (any positive number to the 0 power is 1) and ln(1)= 0 so E(0)= 1 also. Since, for x= 0, the "differ" by 0 that constant is 0: they are the same function.
     
    Last edited: Dec 3, 2011
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