If ln ⁡ x is defined via an integral and e defined from ln ⁡...





If ln x is defined via an integral and e defined from ln x, how would you prove that ln x is the inverse of e x ?
This is a somewhat technically specific question about the relationship between ln x and e x given one possible definition of ln x
Suppose that you define ln x as
ln x = 1 x d t t
ln x = 1 x d t t
We can use the connection between the integral definition of ln x and the harmonic series to show that ln x grows unboundedly. This function is monotonically increasing, so it should have an inverse mapping from its codomain ( R ) to its domain ( R + ). Let's call that function ln 1 x
Since ln x grows unboundedly and is zero when x is 1, we can define e as the unique value such that ln e = 1
So here's my question: given these starting assumptions, how would you prove that ln 1 x = e x (or, equivalently, that ln x = log e x? I'm having a lot of trouble even seeing how you'd get started proving these facts, since basically every calculus fact I know about e and ln x presumes this result to be true.

Answer & Explanation




2022-07-17Added 17 answers

If f ( x ) = log x is defined as a primitive of 1 x for which f ( 1 ) = 0, then f ( a b ) = f ( a ) + f ( b ) holds for any couple of positive real numbers. This gives that the inverse function g ( x ) is a C 1 function, satisfies g ( 0 ) = 1 and the functional equation:
(1) g ( a ) g ( b ) = g ( a + b )
for any couple of real numbers a , b. ( 1 ) and differentiability grants that g = g, hence g ( x ) = e x



2022-07-18Added 3 answers

The advantage of defining ln x as an integral is that it gives a simple way to define e x in elementary calculus (without resorting to infinite series, taking limits of rational powers of e, or other methods).
Once ln x has been defined in this way, you can establish that y = ln x is an increasing function whose range is R , so it has an inverse function with domain R which we can d e f i n e to be y = e x
With this definition, we have the identities
ln ( e x ) = x for all x and e ln x = x for x > 0, and ln e = 1 e 1 = e

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