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Given the axioms:
ln x = definite integral from 1 to x of 1/t dt
ln e = 1

Given the theorems:
ln a^r = r * ln a (for all real numbers r)
ln e^x = x * ln e = x

Prove that e^(ln x) = x

Intuitively, I can see that is true, but how can it be proved?

  • Math -

    ln a^r=r ln a then let a=e
    ln e^x=x ln e but
    ln e=1 so
    ln e^x=x

  • Math -

    In the theorem:

    ln e^x = x * ln e = x

    put x = ln(y):

    ln{exp[ln(y)]} = ln(y)

    Then we have:

    exp(ln(y)) = y

    if it is allowed to conclude from

    ln(a) = ln(b)


    a = b

    This follows from the definition of ln:

    ln x = definite integral from 1 to x of 1/t dt

    Since (for positive x), 1/t in the integrand is always positive, ln(x) is a monotonously increasing function. So, we have that:

    if x > y, then ln(x) > ln(y).

    This is equivalent to saying that

    1) if ln(x) is not larger than ln(y) then x is not larger than y

    Then if

    ln(x) = ln(y)

    then this means that neither is it the case that:

    2) ln(x) > ln(y)

    nor do we have that

    3) ln(y) > ln(x)

    Since 2) is not true, we have by 1):

    4) x is not larger than y

    Since 3) is not true either, we have by 1) (interchange x and y there):

    5) y is not larger than x

    From 4) and 5) it follows that x = y.

  • Math -

    Thanks so much Iblis. Makes perfect sense!

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