${\log}_{a}{x}\neq {\int}^{x}_{1}{\frac{1}{t}}dt$ In most calculus textbooks, $\ln{x}$ is defined to be ${\int}^{x}_{1}{\frac{1}{t}}dt$. Some textbooks validate this definition by demonstrating that this function $\int^{x}_{1}{\frac{1}{t}}dt$ has all the properties of a logarithmic function (I've included pictures of this). I'm skeptical to this particular approach, since we could also define ${\log}_{a}{x}={\int}^{x}_{1}{\frac{1}{t}}dt$. We can still show that the laws of logarithms are properties of this integral, it's also obvious how the algebra will work out. And that means we've justified our claim?

Hell no! The derivative of ${\log}_{a}{x}$ is $\frac{1}{x}{\log}_{a}{e}$. Isn't this approach erroneous then? How then, could we show that this integral does not equal ${\log}_{a}{x}$? We could try showing that some of the properties of the logarithmic functions ($a\neq{e}$) do not hold for ${\int}^{x}_{1}{\frac{1}{t}}dt$. But how do we go about it?

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Your critique is completely correct: the properties shown in those proofs are not enough to distinguish between $\ln x$ and $\log_a x$ for any $a&gt;1$. Indeed, all of the proofs would go through for the function $\int_1^x \frac Ct\,dt$ for any positive constant $C$ as well. (This is secretly the same ambiguity in disguise....)

So yes, you're right that this is not a proof that $\int_1^x \frac 1t\,dt$ must equal $\ln x$ instead of some other $\log_a x = \frac1{\ln a}\ln x$. To be fair, the textbook didn't claim it was such a proof—only that the integral does have logarithm-like properties.

A proof that we really do get $\ln x$ itself, instead of some multiple of it, would need to use some property of the number $e$—which itself depends on what definition of $e$ you choose. One common definition is of all the exponential functions $a^x$ with $a&gt;1$, the number $e$ is the only base $a$ with the property that $\frac d{dx}(b^x)\big|_{x=0} = 1$.

From this one can derive (using the relationship between the derivatives of a function and its inverse function) that $e$ is the only base $a$ for which the inverse function $\log_a x$ satisfies $\frac d{dx} \log_a x\big|_{x=1}=1$. And this additional property is enough to show that $\int_1^x \frac1t\,dt$ is equal to $\log_e x=\ln x$, since the derivative of $\int_1^x \frac1t\,dt$ equals $\frac 1x$ by the fundamental theorem of calculus.
The function $t \mapsto \to \frac 1t$ is continuous at  the interval $ (0,+\infty)$, it has then many antiderivative at $(0,+\infty)$.
The antiderivative  which takes $ 0 $ at $ x=1 $ and $ 1 $ at $ x=e $ is Nepierian logarithm $ (\ln) $.

The one which takes $ 1$ at $x=a$ is $\log_a$.

$$\ln(x)=\log_e(x)=\int_1^x \frac{dt}{t}$$

$${\log_a(x)=\frac{\displaystyle\int_1^x\frac{dt}{t}}{\displaystyle\int_1^a\frac{dt}{t}}}$$