You can, in fact start by defining
ln(x)= \int_1^x \frac{1}{t}dt
From that, because 1/x is continuous for all positive x but not defined for x= 0, we can immediately conclude:
1) ln(x) is defined for all positive x
2) it is continuous and differentiable for all positive x.
3) its derivative is 1/x which is positive for all positive x
4) so ln(x) is an increasing function.
5) ln(1)= \int_1^1 dt/t= 0
6) so ln(x)< 0 for 0<x< 1 and ln(x)> 0 for 1< x.
7) If x is any positive number, then so is 1/x and, by this definition
ln(1/x)= \int_1^{1/x} \frac{1}{t}dt
Let u= xt so that t= u/x and dt= du/x. 1/t= x/u so (1/t)dt= (x/u)(du/x)= du/u.
When t= 1, u= x and when t= 1/x, u= 1 so
ln(1/x)= \int_x^1 \frac{1}{u}du= -\int_1^x\frac{1}{u}du= -ln(x)
8) If x and y are both positive the so is xy and, by this definition,
ln(xy)= \int_1^{xy} \frac{1}{t}dt
Let u= t/x so that t= xu and dt= xdu. Then dt/t= (xdu)/xu= du/u. When t= 1, u= 1/x and when t= xy, u= y so
ln(xy)= \int_{1/x}^y \frac{1}{u}du= \int_{1/x}^1 \frac{1}{u}du+ \int_1^y \frac{1}{u}du
= -\int_1^{1/x} \frac{1}{u}du+ \int_1^y \frac{1}{u}du= -ln(1/x)+ ln(y)
= -(-ln(x))+ ln(y)= ln(x)+ ln(y)
9) If x is a positive real number and y is any real number, then x^y is a positive real number and, by this definition,
ln(x^y)= \int_1^{x^y} \frac{1}{t}dt
If y is not 0, let u= t^{1/y} so that t= u^y and dt= y u^{y-1}du and dt/t= yu^{y-1}du/u^y= y du/u. When t= 1 u= 1 and when t= x^y u= x so
ln(y^x)= y\int_1^x \frac{1}{u}du= y ln(x)[/itex].<br />
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If y= 0, then x^y= x^0= 1 so ln(x^y)= ln(1)= 0= 0 ln(x) so that "ln(y^x)= y ln(x) for any real y.<br />
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10) Since ln(x) is continuous and differentiable for all positive real numbers, we can apply the <b>mean value theorem</b> to any interval of positive real number and, in particular to the interval [1, 2]. <br />
\frac{ln(2)- ln(1)}{2- 1}= \frac{ln(2)- 0}{1}= ln(2) <br />
and the mean value theorem says there must be some number, c, between 0 and 1 such that ln(2) is equal to the derivative at c: ln(2)= 1/c. Since c< 2, 1/c> 1/2 and it follows that ln(2)> 1/2. That is important because we can say that, for any positive real number, X, ln(2^{2X})= 2X ln(2)&gt; 2X(1/2)= X. That is, given any positive real number, there exist x such that ln(x) is larger- ln(x) is NOT bounded above.<br />
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Since ln(x) is an increasing function, it follows that \lim_{x\to \infty} ln(x)= \infty. Also, since ln(1/x)= -ln(x), \lim_{x\to 0}ln(x)= \lim_{y\to \infty} ln(1/y)= -\lim_{y\to\inty} ln(y)= -\infty. That is, ln(x) is a one-to-one function mapping the set of all positive real numbers <b>onto</b> the set of all real numbers.<br />
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Therefore, ln(x) has an inverse function, which I will call "Exp(x)" from the set of all real numbers to the set of all positive real numbers. All of the properties of "Exp(x)", such as the fact that the derivative of Exp(x) is again Exp(x), that Exp(x+ y)= Exp(x)Exp(y), etc. can be derived from the corresponding ln laws above but the crucial point, answering your question, is this:<br />
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If y= Exp(x), then, by definition of "inverse" function, x= ln(y). <b>If x is not 0</b>, we can divide both sides of the equation by it: 1= (1/x)ln(y)= ln(y^{1/x}). Now go back to the "exponential" form: y^{1/x}= Exp(1) from which it follows that y= (Exp(1))^x. If x= 0, then y= Exp(0)= 1 because ln(1)= 0 and y= 1= Exp(1)^0= Exp(1)^x. That is, for all x, Exp(x)= Exp(1)^x so that Exp(x), the inverse of ln(x), really is an exponential function. We can, of course, define e= Exp(1) to get that the inverse to ln(x) is e^x