It's better to define as the inverse function of . Since the range of is , that means the domain of is
It follows therefore, that
It follows then that if , then .
Therefore your definition follows from the Fundamental Theorem of Calculus.
It's better to define as the inverse function of . Since the range of is , that means the domain of is
It follows therefore, that
It follows then that if , then .
Therefore your definition follows from the Fundamental Theorem of Calculus.
In mathematics, one does not need a justification for definitions. One is free to define whatever they like as long as the objects are indeed "well-defined." (This may warrant a separate discussion.) The only possible flaw with a definition is that other mathematicians may not find it interesting and may, e.g., refuse to accept a conference paper about it, but this is not a properly mathematical issue.
I am not sure why in this case the authors decided to define logarithm this way. Maybe they found that proving properties of logarithm is easier this way.
Yes, ln(x) can be defined as the inverse of but I, personally, don't agree with Prove It that that definition is "better".
For one thing, we can see immediately from this definition that ln(x) is differentiable for all for all positive x and the derivative is 1/x while finding the derivative of involves showing that which, while possible, is not trivial.
We can prove a number of the basic properties of the logarithm directly from that definition:
Since its derivative, 1/x, is positive for all x, ln(x) is an increasing function.
To prove ln(1/x)= -ln(x), let u= 1/t in . From u= 1/t, t= 1/u so ,when t= 1, u= 1, and when t= 1/x, u= x so the integral becomes .
To prove ln(xy)= ln(x)+ ln(y), let u= t/y so t=yu dt= ydu, when t=1, u= 1/y, and when t= xy, u=x so the integral becomes and since ln(1/y)= -ln y, that is, ln(x)+ ln(y),
To prove [tex]ln(a^y)= y ln(a), if , let so that and . When t= 1, u= 1 and when , u= x so the integral becomes .
(If y= 0, then and so
Since ln(x) is differentiable for all x, we can apply the mean value theorem to any interval of real numbers and, in particular, to [1, 2]. The mean value theorem says that there exist c between 1 and 2 so that the mean value theorem says (ln(2)- 1)/(2- 1)= (ln(2)- 0)/1= ln(2)= 1/c. Since , . That is, ln(2)> 1/2. From that, for any X> 0, . That means that the ln(x) function has no upper bound and, since 1/x is positive for all positive x, . Since ln(1/x)= -ln(x), . That is, ln(x) maps the set of all positive real numbers, one to one, to the set of all real numbers.
That tells us that ln(x) has an inverse which we can call "exp(x)" (I am deliberately avoiding calling that function ) which maps all real numbers to the positive real numbers. And, of course, since the derivative of ln(x) is 1/x, it follows that if y= exp(x), then x= ln(y) so that dx/dy= 1/y and thus dy/dx= y= exp(x).
But I think the main thing you want is this: If y= exp(x) then x= ln(y). If , . Therefore and so . If x= 0, 0= ln(y) means that y= 1= exp(1)^0. In either case, this function, exp(x), really is a some number to the x power. If we define e to be exp(1), the value of x so that ln(x)= 1, we have .