If we write $\ln(x):=\int_{1}^{x}\frac{1}{t}dt$, can we show that the natural logarithm tells us the power $e$ must be raised to in order to get $x$? The way I was first introduced to the logarithm was that $\log_ab$ means 'the power you need to raise $a$ by in order to get $b$'. Additionally, $a$ and $b$ must be greater than $0$. If I tried to formalise this notion, then I would probably come up with something like
$$
\log_ab := n \text{ such that } a^n=b, \text{ where } a,b>0
$$
However, I have found that many logarithms are instead defined using integrals. In particular, the natural logarithm is often defined as
$$
\ln(x):= \int_{1}^{x}\frac{1}{t}dt
$$
Is there a way of showing that this definition still satisfies my intuitive conception of what a logarithm is? In other words, can we show that
$$
\ln(x):= \int_{1}^{x}\frac{1}{t}dt \implies \ln(x)=n \text{ such that } e^n=x, \text{ where } x>0
$$
Even better would be to show that
$$
\ln(x)= \int_{1}^{x}\frac{1}{t}dt \iff \ln(x)=n \text{ such that } e^n=x, \text{ where } x>0
$$
Where you use either the LHS or the RHS as the starting point for your definition.
 A: Both the functions $\exp(x)$ and $\ln(x)$ are strange in that they can be defined in a variety of ways, all of which are equivalent. You can, for example, define $\exp(x)$ to be either of the following:
$$
\exp(x) = \lim_{n \to \infty} \left(1 + \frac 1 n\right)^{nx}, \quad \textrm{or} \quad  \exp(x) = \sum_{k=0}^\infty \frac{x^k}{k!}.
$$
Yet another definition gives $\exp(x)$ as the unique function $f(x)$ with the property that $f'(x) = f(x)$ and $f(0) = 1$ (and yes, $\exp(x)$ is the only function with this property). In each of these instances, it is a different (but not terribly difficult) matter to show that $\exp(x+y) = \exp(x)\exp(y)$ from these definitions. So really what you want to do is prove that using one of these definitions, $\exp(\ln(x)) = x$ and $\ln(\exp(x)) = x$ for $x > 0$.
Let's suppose we take the approach that $\exp(x) = \exp'(x)$ and $\exp(0) = 1$, for whatever this function $\exp(x)$ actually is, and $\ln(x) = \int_1^x \frac 1 t \,dt$. Let $g(x) = \ln(\exp(x))$. Then $g(0) = 0$, and since $\frac{d}{dx} \ln(x) = \frac 1 x$ by the fundamental theorem of calculus,
$$
g'(x) = \frac{\exp'(x)}{\exp(x)} = \frac{\exp(x)}{\exp(x)} = 1
$$
from which we can deduce that $g(x) = x$. A similar strategy can be used to show $\exp(\ln(x)) = x$, so $\exp$ and $\ln$ are function inverses.
A: Yes: in this approach, $\mathrm e$ is the number such that
$$\ln \mathrm e=\int_1^{\mathrm e}\frac{\mathrm d t}{t}=1.$$
Therefore,
$$\mathrm e^{y}=x\iff y\ln\mathrm e=y=\ln x. $$
A: Note that the function $\ln$ defined in the question is one-to-one.
Assume that $e$ is such that $e^x$ is the unique function $y$ that satisfies $y’=y,y(0)=1$. Then the question boils down to showing that the following function $y=f(x)$ equals $x$:
Proposition. If $y=f(x)=e^{\int_1^x 1/t~dt},f(1)=1$, then $f(x)=x.$
Proof. Taking derivatives, one has $$\frac{dy}{dx}=\frac yx$$ $$\Rightarrow \frac 1 y~dy=\frac 1 x~dx$$
$$\Rightarrow \ln y=\ln x,~{\rm since~}f(1)=1$$
$$\Rightarrow y=x,$$ since $\ln$ is one-to-one.
