Need help in making sense of natural logarithms and exponents in terms of calculus I understand that the natural logarithm function (inverse of the exponential function) is defined to be:
$ln(x) = $$\int_1^x 1/t\,dt$$ $
Which I understand to be a function which gives the area under the graph of $1/t$ from 1 to x. Also, it is a function whose instantaneous rate of growth from 1 to $x$ is $1/t$ (according to what I understand of anti-derivatives and integrals from calculus).
I understand that $x$ and $t$ mean different things here.
The definition of $ln(x)$ in terms of Euler's number is $log_e x$. That is, the function yields the value $e$ should be raised to, to get $x$.
What I'd like to understand better is how both these definitions work together.
Essentially we have:
$\int_1^x 1/t\,dt$ = $log_e x$ = The value $e$ should be raised to to get $x$.
I'd like to understand better how the function $f(t) = 1/t$ relates to $e$. From the definitions above, I surmise it is correct to say that:
"Raising $e$ to a value equivalent to the area from 1 to $x$ under the graph of function $f(t) = 1/t$ yields $x$"
or
"$e$ raised to a value equivalent to a function whose rate of growth from 1 to $x$ is $1/t$, yields $x$"
Are these statements correct? How does the function from the second definition relate to the definition of $e^x$ being a function whose rate of growth is equal to itself?
 A: We can show that if $\ln x$ is a function that "yields the value $\mathrm e$ should be raised to to get $x$", in your words, is one whose derivative is $\frac1x$. The algebraic definition I quoted looks like this in a more mathematical language:
$$\ln (\mathrm e^y)=y.$$
Here, $\mathrm e^y=x$. If we apply the exponential function to both sides, we also get
$$\begin{align}\mathrm e^{\ln \mathrm e^y}&=\mathrm e^y\\
\mathrm e^{\ln x}&=x.\end{align}$$
In this form, we can apply a formal trick: take the derivative on both sides to get
$$\ln'(x)\mathrm e^{\ln x}=1.$$
Divide by $\mathrm e^{\ln x}$ to get
$$\ln'(x)=\frac{1}{\mathrm e^{\ln x}}=\frac{1}{x}.$$
So we found the derivative of $\ln x$. What we needed to get there is firstly that $\ln \mathrm e^x=x$, and secondly that the derivative of the exponential function is again the exponential function. By the way, I said formal trick because technically speaking, we didn't know that $\ln$ even has a derivative, so applying the chain rule to a function containing it is a bit iffy. But there are rigorous proofs which show that such a thing is justified if $\mathrm e^x$ has a derivative, which we know is true.
So now by the fundamental theorem of calculus, we see that $\ln x$ can potentially be written as an integral function of $\frac{1}{x}$, since it's an antiderivative of $\frac{1}{x}$. We just have to find a fitting lower boundary for the integral so that the constant of integration is the right one. Since $\ln 1=\ln\mathrm e^0=0$, we should choose $1$ as the lower boundary, since then
$$\int_1^x\frac{1}{t}\mathrm dt =\ln x-\ln 1=\ln x-0=\ln x.$$
And thus we found an integral representation of the natural logarithm by using our knowledge of how it interacts with the exponential function, whose derivative we know, and the fundamental theorem of calculus. And your statements are found to be correct.
