Develop an elegant approach to the exponential and logarithm functions. Define a function G(x) f

Develop an elegant approach to the exponential and logarithm...

Key Concepts

Integral Definition of the Logarithm

This approach defines the logarithm function as an integral, specifically by G(x) = ??? 1/t dt for x > 0. This definition allows one to derive all properties of the logarithm without assuming any prior knowledge of the exponential or logarithm functions, emphasizing a constructive and analytical foundation.

Substitution in Integration

The substitution method is used to change the variable of integration, which simplifies the process of evaluating an integral or proving integral identities. In the context of this problem, it is used to demonstrate invariance under scaling, for example showing that ???? 1/t dt equals ??? 1/t dt by substituting u = t/a.

Additive Property of the Logarithm

One fundamental property derived from the integral definition is that G(ab) = G(a) + G(b), mirroring the logarithmic identity ln(ab) = ln(a) + ln(b). This is established by splitting the interval of integration and then applying the substitution, demonstrating the deep link between multiplication and addition in logarithms.

Power Laws for Logarithmic Functions

The properties G(a?) = n G(a) and G(a^(1/n)) = (1/n) G(a) for integers n, and more generally G(a?) = r G(a) for rational r, encapsulate the power rules of logarithms. These properties highlight how exponentiation in the domain translates to multiplication outside the function, underpinning many applications of logarithms in algebra.

Fundamental Theorem of Calculus

The FTC connects the concept of integration and differentiation. In the context of this approach, it is used to conclude that G(x) is continuously differentiable with derivative 1/x, thereby not only confirming its smooth behavior but also establishing its monotonic nature.

Monotonicity and Continuity

Proving that G(x) is increasing relies on the fact that its derivative is positive for all x > 0. This property, stemming from the fundamental theorem of calculus, ensures that G(x) continuously and uniquely maps the domain (0, infinity) to the real numbers, an essential characteristic for defining an inverse function.

Existence and Uniqueness via the Intermediate Value Theorem

The problem demonstrates that there exists a unique number E such that G(E) = 1. This is an application of the Intermediate Value Theorem, which guarantees that a continuous and strictly monotonic function attains every value exactly once within its range. This uniqueness is crucial as it establishes a natural base for the exponential function.

Transcript

00:01 Okay, so we're giving that t of x is equal to this integral of x or 1 2x of 1 over t d t and that x is graded in z.

00:12 Okay, for part a, we're asked to show that a, b, 1 over t, d, t is equal to integral from 1 over b of 1 over t, d t for all.

00:25 A, b, greater than so.

00:27 We're asked to use u is equal to t over a, but then d u is equal to 1 over a d t.

00:35 So we have a d u equal to d t.

00:41 Okay, let's look at our left hand side.

00:44 Work with that.

00:46 Okay, so we said d t is a d u, and then what is t? well, t is this a u, it's t.

00:56 So we have a u, or a is cancel.

01:00 And let's rewrite our endpoints.

01:04 So we have u is equal to ab over a, that just gives me b, and u is equal to a over a.

01:10 That gives me one, so we have b, 1 to b.

01:13 That's equal to 1 to 1 to b of 1 over u, d, u.

01:20 And this actually looks like that.

01:22 If you take the integral of it, you know that they are the same.

01:26 Okay, now automatically part b.

01:27 We have to show that g of a, b equal to g of a plus g of b and to use part a.

01:39 Okay, well, g of a, b is equal to the integral from 1 to a, b of 1 over t, b, of 1 over t, d, t.

01:48 Well, that's equal to, let's break that up.

01:51 Let's break it up at point a, because we know a and b are graded in it equal to 0.

02:01 Or they're actually graded in jail.

02:04 So if i do a times b, or actually if i have a b, then a must be less than a times b.

02:16 Okay, so we have one to a, one over t, d, t.

02:20 And this is actually already g of a plus integral from a, b, a of 1 over t, d, t.

02:29 And based on part one, this is basically 1 to b of 1 over t.

02:34 D t plus this so this is g a plus gb okay now for part c let's see what we're asked we're asked to show that g of 1 is equal to 0 and that g of a inverse is equal to negative g a for all a equated in so okay so we have from integral from 1 to 1 to 1 over t d t that's equal to ln of t, evaluated at 1, 1, or ln of 1 minus ln of 1.

03:20 These two cancel, so we just get 0.

03:23 Okay, so this works.

03:25 Now we're asked to show this, so g, a, to the fire of negative 1 is a, 1, 1 over t, d, t.

03:36 Okay, that's equal to ln of t evaluated at a, negative 1, and 1.

03:41 That gives me ln of a negative 1 minus ln of 1.

03:49 Ln of 1 is 0 0 .1 is 0 0 .0 .1 is 0 .00 is 0 .00.

03:57 We can pull out a power, bring it to the front.

04:00 We get negative 1, ln of a.

04:05 Okay.

04:06 Now if we take the integral, or the negative integral from 1 to a of 1 over t, this is equal to negative g .a.

04:18 That's equal to negative ln of g evaluated at a and 1, negative ln of a minus plus ln of 0, or 1, which is 0.

04:28 So we get negative ln of a is equal to negative ln of a.

04:32 So g, a to the power of negative 1 is equal to negative g .a.

04:38 Okay, now for part d, we're asked to show that ga to equal to n, to the power of n for all a graded in zero and n contained in integers okay so g of a to is equal to the integral from a to n one of one over t d t that's equal to ln of t evaluated at a and one that gives me ln of a to the power of n minus ln of one this is zero you can pull out of power you get n ln of a now n g a sorry this there's no power of n here okay that's equal to n one over t d t from a from one to a that gives me n ln of t evaluated at a and one that gives me n ln of t evaluated at a and one that gives me n ln of a minus n ln of one this cancels to zero so where this left with n l n of a so these two are equal so n g a is equal to g a to the power of n okay for parts e we're asked to show that g a to 5 over n is equal to 1 over n g a a greater than zero and n not equal to zero.

06:27 Okay.

06:28 So g, a to the power of 1 over n, equal to the integral, a, 1 over n, 1, 1, 2, d, t.

06:37 That gives me ln of t, evaluated at a to the power of 1 over n, and 1, that gives me ln.

06:45 Okay, so just to state, i'm just going to not write ln of 1 anymore, because every time we have this or ln of something minus ln of 1, but that's this there also is left with ln of a...

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