Prove withoui calculus: the integral of 1/x is logarithmic It was known in the 17th century that the function
$$ t \mapsto \int_{1}^{t} \frac{dx}{x} $$
is logarithmic: a geometric sequence in the domain produces an arithmetic sequence in the codomain. This is. of  course, easy to prove with the fundamental theorem of calculus.
But is there a simpler, perhaps geometric, way of proving this?
 A: Consider this integral and the substitution $tx=y$:$$\int_1^u \frac{1}{x}dx=\int_t^{tu}\frac{t}{y}\frac{dy}{t}=\int_t^{tu}\frac{dy}{y}$$
From this follows:
$$\int_1^t \frac{dx}{x}+\int_1^u \frac{dx}{x}=\int_1^{tu} \frac{dx}{x}$$
The geometric way of describing this is stretching the function $f(x)=\frac{1}{x}$ with a horizontal factor $t$, so it becomes $f(x)=\frac{1}{x/t}$. Then we divide this function with the same factor $t$.
A: It has to do with the geometry of the hyperbola. Specifically, that the function $x \mapsto \frac{1}{x}$ maps geometric sequences to geometric sequences.
To demonstrate, let A denote the region above the abscissa and below the hyperbola $yx = 1$ and between $x = 1$ and $x = 2$. And similarly let B denote the region above the abscissa and below the hyperbola and between $x = 2$ and $x =4$. 
Then A is the image of B under the following transformation:
$$(x,y) \mapsto \left( \frac{x}{2}, 2x \right).$$
This can be checked by transforming the boundaries of B.
This transformation obviously preserves areas, so therefore A and B have equal areas. Or, in formula:
$$\int_{1}^{2}\frac{dx}{x} = \int_{2}^{4} \frac{dx}{x}.$$ 
Defining the natural logarithm as $\int_{1}^{t}\frac{dx}{x}$ means that the above equation can be written as $\log 4 = 2 \log 2$. A similar argument can prove that, more generally, $\log st = \log s + \log t$. 
