Math
posted by Sean .
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?

Given
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 
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)
that
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. 
Thanks so much Iblis. Makes perfect sense!