Proof that $\int_1^x \frac{1}{t} dt$ is $\ln(x)$ A logarithm of base b for x is defined as the number u such that $b^u=x$. Thus, the logarithm with base $e$ gives us a $u$ such that $e^u=b$. 
In the presentations that I have come across, the author starts with the fundamental property $f(xy) = f(x)+f(y)$ and goes on to construct the natural logarithm as $\ln(x) = \int_1^x \frac{1}{t} dt$. 
It would be suprising if these two definitions ended up the same, as is the case. How do we know that the are? The best that I can think of is that they share property $f(xy) = f(x)+f(y)$, and coincide at certain obvious values (x=0, x=1). This seems weak. Is their a proof?
 A: If you want the logarithm to be expressed as a function $f$ then the most important properties have to hold, which are
$$\log(xy) =\log(x)+\log(y)$$
$$\log(x^a) =a\log(x)$$
$$f(1)=0$$
And... suppose $f$ admits a derivative. Then fixing $y$ and differentiating the first equation gives:
$$yf'(xy) =f'(x)$$
Putting $x=1$ gives
$$f'(y) =\dfrac{f'(1)}{y}$$ for each $y\neq0$
From this equation we se the derivative is monotonous en each interval not containing the origin. Morover, $f'$ is continuous in $(c,x)$ with $c>0$ so we can apply the second $\mathcal{FTC}$:
$$f(x) - f(c) = \int_c^x f'(t) dt =f'(1) \int_c^x\frac{1}{t}dt$$
If $x>0$ this equation is valid for each nonnegative $c$, so choosing $c=1$ gives:
$$f(x) = f'(1) \int_1^x \frac{dt}{t}$$
You can readily check from this equation that the previous properties are met. Moreover, you can check that the logarithm will be the unique function that will satisfy the above requierements and $f'(1)=1$, which will give you the desired definition:
$$\log x = \int_1^x \frac{dt}{t}$$
What I gave you is rather a stub from Apostol's Calculus, pages 278-281, 2nd ed.
A: If x,y>0, Define $x=e^u,y=e^v$,so $f(e^{u+v})=f(e^u)+f(e^v)$,Define $g(u)=f(e^u)$,then
$$g(u+v)=g(u)+g(v)$$
this is Cauchy Functional Equation ,g(u)=cu,
the proof you can see How to prove $f(x)=ax$ if $f(x+y)=f(x)+f(y)$ and $f$ is locally integrable
I think you can finish the following.
A: The following properties uniquely determine the natural log:
1) $f(1) = 0$.
2) $f$ is continuous and differentiable on $(0, \infty)$ with $f'(x) = \frac{1}{x}$.
3) $f(xy) = f(x) + f(y)$

We will show that the function $f(x) = \int_1^x \frac{1}{t} dt$ obeys properties 
1,2, and 3, and is thus the natural log.
1) This is easy, since $f(1) = \int_1^1 \frac{1}{t} dt = 0$.
2) Defining $f(x) = \int_1^x \frac{1}{t} dt$, we note that since $\frac{1}{t}$ is continuous on any interval of the form $[a,b]$, where $0 < a \leq b$, then the Fundamental Theorem of Calculus tells us that $f(x)$ is (continuous and) differentiable with $f'(x) = \frac{1}{x}$ for all $x \in [a,b]$.
3) 
$$\begin{align}
f(xy) = \int_1^{xy} \frac{1}{t}dt &= \int_1^x \frac{1}{t} dt + \int_x^{xy} \frac{1}{t} dt
\\
&= f(x) + \int_{1}^{y} \frac{1}{u} du
\\
&= f(x) + f(y)
\end{align}$$
where in the last step we perform the substitution $t = ux$ (viewing $x$ as constant).  
A: Put $l(x) = \int_1^x dt/t.$  Then $l'(x) = 1/x$ if $x > 0$ by the fundamental theorem of calculus.  Since $l' > 0$ on $(0,\infty)$, $l$ is 1-1 and therefore has an inverse.  Denote its inverse by $m$.  We have $l(m(x))= x;$ differentiate to see that
$$1 = l'(m(x))m'(x) = m'(x)/m(x).$$
Multiply to get $m'(x) = m(x).$  It is not at all hard to see that $m(0) = 1.$ Therefore, $m(x) = e^x$.
A: Let $f(x) = e^{\int_1^x \frac{1}{t} dt}$.
Then $f'(x)=\frac{1}{x}e^{\int_1^x \frac{1}{t} dt}=\frac{f(x)}{x}$.
hence
$$x f'(x)-f(x)=0 \Rightarrow \frac{x f'(x)-f(x)}{x^2}=0 \Rightarrow \left(\frac{f(x)}{x}\right)'=0 \,.$$
A: I've done this before, but here I go again.
If $f(xy) = f(x)+f(y)$, 
$f(1)=0$
and
$f(x+h)-f(x) = f\left(x\left(1+\dfrac{h}{x}\right)\right)-f(x)$
$= f(x)+f\left(1+\dfrac{h}{x}\right)-f(x)$
$= f\left(1+\dfrac{h}{x}-f(1)\right)-$
so 
$\dfrac{f(x+h)-f(x)}{h} = \dfrac{f\left(1+\dfrac{h}{x}\right)-f(1)}{h}
= \dfrac{1}{x} \dfrac{f\left(1+\dfrac{h}{x}\right)-f(1)}{\dfrac{h}{x}}
$.
Letting $h \to 0$,
$f'(x) = \dfrac{f'(1)}{x}$.
Choose $f'(1) = 1$ to get the natural log.
Similarly, starting with
$f(x+y) = f(x)f(y)$
we get
$f'(x) = f'(0)f(x)$.
