My book defines this function in the provided attachment. My confusion is why is this true? No where in my book does it explain it with scarcely any adequacy.
It's better to define $\displaystyle \displaystyle \begin{align*} \ln{x} \end{align*}$ as the inverse function of $\displaystyle \displaystyle \begin{align*} e^x \end{align*}$. Since the range of $\displaystyle \displaystyle \begin{align*} e^x \end{align*}$ is $\displaystyle \displaystyle \begin{align*} (0, \infty) \end{align*}$, that means the domain of $\displaystyle \displaystyle \begin{align*} \ln{x} \end{align*}$ is $\displaystyle \displaystyle \begin{align*} (0, \infty) \end{align*}$
It follows therefore, that
$\displaystyle \displaystyle \begin{align*} y &= \ln{x} \\ e^y &= x \\ \frac{d}{dx}\left(e^y\right) &= \frac{d}{dx}(x) \\ \frac{d}{dy}\left(e^y\right)\frac{dy}{dx} &= 1 \\ e^y\,\frac{dy}{dx} &= 1 \\ \frac{dy}{dx} &= \frac{1}{e^y} \\ \frac{dy}{dx} &= \frac{1}{x} \end{align*}$
It follows then that if $\displaystyle \displaystyle \begin{align*} x > 0 \end{align*}$, then $\displaystyle \displaystyle \begin{align*} \int{\frac{1}{x}\,dx} = \ln{x} + C \end{align*}$.
Therefore your definition follows from the Fundamental Theorem of Calculus.
Of course they're not inverse functions, what you have graphed are the exact same function, as $\displaystyle \displaystyle \begin{align*} y = \ln{x} \end{align*}$ is equivalent to $\displaystyle \displaystyle \begin{align*} x = e^y \end{align*}$.
What you should be graphing are $\displaystyle \displaystyle \begin{align*} y = e^x \end{align*}$ and $\displaystyle \displaystyle \begin{align*} y = \ln{x} \end{align*}$.
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 $\displaystyle e^x$ 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 $\displaystyle e^x$ involves showing that $\displaystyle \lim_{h\to 0}\frac{e^h- 1}{h}= 1$ 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 $\displaystyle \int_1^{1/x} dt/t$. From u= 1/t, t= 1/u so $\displaystyle dt= -u^{-2}du$,when t= 1, u= 1, and when t= 1/x, u= x so the integral becomes $\displaystyle \int_1^x (1/(1/u))(-(1/u^2)du)= -int_1^x (1/u)du= -ln(x)$.
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 $\displaystyle \int_{1/y}^x (1/yu)(ydu)= \int_{1/y}^x (1/u)du= \int_{1/y}^1 du/u+ \int_1^x du/u= -\int_1^{1/y} du/u+ \int_1^x du/u= -ln(1/y)+ ln(x)$ and since ln(1/y)= -ln y, that is, ln(x)+ ln(y),
To prove [tex]ln(a^y)= y ln(a), if $\displaystyle y\ne 0$, let $\displaystyle u= t^{1/y}$ so that $\displaystyle t= u^y$ and $\displaystyle dt= y u^{y- 1}du$. When t= 1, u= 1 and when $\displaystyle t= x^y$, u= x so the integral becomes $\displaystyle \int_1^x (1/u^y)(y u^{y-1}du= y\int_1^x du/u= yln(x)$.
(If y= 0, then $\displaystyle x^y= x^0= 1$ and so $\displaystyle ln(x^y)= ln(1)= 0= 0(ln(x))= yln(x)$
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 $\displaystyle 1\le c\le 2$, $\displaystyle 1/2< c< 1$. That is, ln(2)> 1/2. From that, for any X> 0, $\displaystyle ln(2^{2X})= 2Xln(2)> X$. That means that the ln(x) function has no upper bound and, since 1/x is positive for all positive x, $\displaystyle \lim_{x\to\infty} ln(x)= +\infty$. Since ln(1/x)= -ln(x), $\displaystyle lim_{x\to 0} ln(x)= -\infty$. 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 $\displaystyle e^x$) 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 $\displaystyle x\ne 0$, $\displaystyle 1= (1/x)ln(y)= ln(y^{1/x})$. Therefore $\displaystyle y^{1/x}= exp(1)$ and so $\displaystyle y= exp(x)= (exp(1))^y$. 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 $\displaystyle exp(x)= e^x$.