# Math

posted by on .

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 - ,

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

• 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)

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.

• Math - ,

Thanks so much Iblis. Makes perfect sense!