Prove $\ln(a \cdot b)=\ln(a)+\ln(b)$, $a,b>0 $ and $\lim_{a \to 0} \frac{\ln(1+a)}{a} =1$.But we shall not even know that the exponential function. In Japanese high school education, the number $e$ is defined first. There is no disagreement that $e^q$ can be defined algebraically when the rational number $q$ is an exponent. However,  there is a sudden "deregulation" that $q$ can be a real number. Until now, I had no doubts about this. But now I'm confused about it.
(I am a user of college-level mathematics, and probably know the epsilon-delta argument. But I'm not familiar with rigorous mathematics.)
I found the way exponential functions were introduced in Japanese high school to be a bit mathematically imprecise. I overheard someone say, "That can be essentially eliminated with only high school knowledge by starting with defining the logarithmic function in terms of integrals. My question is based on that sense. Specifically;

*

*We assume that we do not know the existence or properties of the $\exp(x)$.

*We assume that we do not know that $\ln(x)$ is the logarithm.

*But, we assume that we know general properties of differentiation and integration, such as integration by substitution.

*We also assume that the function $\ln$ is defined as follows;
$$\ln(x):=\int_{t=1}^{t=x} \frac{1}{t}\ dt \tag{0-1}$$

*We do not know any more about ln. For example, we do not know the following formulas;
$$\lim_{a \to 0}\ \frac{\ln(1+a)}{a} =1 \tag{0-2}$$
My Question:

Under the these assumptions, I would like to derive the following formulas. In particular, we want to pay attention to whether we are violating mathematical rules (e.g., improper exchange of integrals and limits).
$$\ln(a \cdot b)=\ln(a)+\ln(b) \ , a,b>0 \tag{1}$$
$$\lim_{a \to 0} \frac{\ln(1+a)}{a} =1 \ ,a>0 \tag{2}$$

Perhaps an outline of the solution would be as follows;
For (1)
$$\ln(a \cdot b):=\int_{t=1}^{t=a \cdot b} \frac{1}{t} dt$$
$$=\int_{t=1}^{t=a} \frac{1}{t}\ dt + \int_{t=a}^{t=a \cdot b} \frac{1}{t}\ dt$$
$$=\ln(a) + \int_{t=a}^{t=a \cdot b} \frac{1}{t}\ dt \tag{1-1}$$
Using $t=s*a$, second term on the right side is
$$\int_{t=a}^{t=a \cdot b} \frac{1}{t}\ dt
=\int_{s=1}^{s=b} \frac{1}{sa} \cdot a\ ds
=\int_{s=1}^{s=b} \frac{1}{s} \ ds
=\ln(b) \tag{1-2}$$
For (2)
$$\ln(1)=\int_{t=1}^{t=1} \frac{1}{t}\ dt=0 $$
Therefore,
$$\frac{\ln(1+a)}{a}
=\frac{\ln(1+a)+0}{a}$$
$$=\frac{\ln(1+a)-\ln(1)}{a}  + \frac{\ln(1)}{a}  
=\frac{\ln(1+a)-\ln(1)}{a}  +0 $$
If we take the limits of both sides of the above equation, we can show the (2).
That is;
$${\lim}_{a \to 0} \frac{\ln(1+a)}{a}=
{\lim}_{a \to 0}\frac{\ln(1+a)-\ln(1)}{a} 
=\ln'(1)=\frac{1}{1}=1$$
Again, I emphasize that my main concern is not the solution itself, but the  "violate mathematical rules". In other words, does the above solution do anything mathematically inappropriate? (e.g., use of circular argument, improper ordering of integrals and limits, etc.)
 A: It's really good that you are noting this must be proven and not taken for granted!
Integration by substitution is valid.  So your proof is valid.
Note if you use Reimann definition of integral
$\int_a^b f(x) dx = \lim_{n\to\infty} \sum_{i=1}^nf(x_i)\Delta x$ where we divide the interval $[a,b]$ into $n$ equal parts, $\Delta x = \frac {b-a}n$ and $x_i \in $ the $i$th of the smaller intervals (for practical purposes we may assume, assuming the function is integrible at all, the endpoint).
In other words: $\int_a^b f(x)dx = \lim_{n\to\infty}\sum_{i=1}^n f(a + i\frac {b-a}n)\cdot \frac {b-a}n$.
So
$\int_{a}^{b} \frac 1x dx = \lim_{n\to\infty} \sum_{i=1}^n\frac 1{x_i}\Delta_1 x=$ (where $\Delta_1 x = \frac {b-a}n$ and $x_i = a + i\Delta_1 x$.  Now let $k$ be any non-zero constant....)
$\lim_{n\to\infty} \sum_{i=1}^n\frac 1{kx_i}k\Delta_1 x= $(where $k\Delta_1 x = \frac {kb- ka}n$ and $kx_i = ka + i(k\Delta_1 x)$)
$\lim_{n\to\infty} \sum_{i=1}^n\frac 1{y_i}\Delta_2 y= $(where $\Delta_2 y = k\Delta_1 x = \frac {kb-ka}n$ and $y_i = ka + i\Delta_2 y$)
$\int_{ka}^{kb} \frac 1y dy= \int_{ka}^{kb} \frac 1x dx$
So for any non-zero constant, $k$ we have $\int_a^b \frac 1x dx = \int_{ka}^{kb} \frac 1x dx$.
(we can give a similar argument for the other definitions of integral)
And with that....
A: Here's a different proof of the formula $\ln(ab)= \ln(a) + \ln(b)$.
Suppose that $a$ is a positive constant.
And suppose $f$ is the function defined over the positive reals by:
$f(x) = \ln(ax) - \ln(x) - \ln(a)$.
$f$ is differentiable over the positive reals. By the chain rule,
$f'(x) = a \frac{1}{ax} - \frac1{x} = 0$
So $f$ is constant.
As such, for all positive reals $x$, $f(x) = f(1) = \ln(a) - \ln(a) -\ln(1) = 0$.
Which concludes our proof.
