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!
To prove that e^(ln x) = x using the given axioms and theorems, we can follow these steps:
Step 1: Start with the expression e^(ln x).
Step 2: Apply the theorem ln a^r = r * ln a by setting a = e and r = ln x. This gives us e^(ln x) = x * ln e.
Step 3: Use the given axiom ln e = 1 to simplify the expression on the right-hand side. We now have e^(ln x) = x * 1.
Step 4: Simplify the right-hand side of the equation to give e^(ln x) = x.
Step 5: Since the left-hand side and right-hand side of the equation are equal, we have proved that e^(ln x) = x.
It's important to note that in this proof, we used the properties of logarithms and exponentials, specifically the relationship between the natural logarithm and the exponential function. These properties allow us to simplify the expression and arrive at the desired result.