This value would be the solution to:$$\int_{0}^{a}\ln(x)dx=1$$ With integration by parts this is well known to be $[x\ln(x)-x]^a_0=1$ or what is the same:$$a\ln(a)-a=1$$ The limit at $0$ easily dismissed as $0$ with L'Hôpital.
Oddly, WolframAlpha outputs $e^{W(1/e)+1}$ as a solution, where $W(z)$ is the product log function. This is aproximately $3.6$. But if I input $$\int_0^{e^{W(1/e)}+1}\ln(x)dx$$ it doesn't compute. Surely I must have gone wrong somewhere?
On
Here is $$ \int_0^a\ln(x)\;dx $$
The spot where it crosses the line $y=1$ is, indeed, $$ e^{W(1/e)+1}=\frac{1}{W(1/e)} \approx 3.5911 $$
This integral is a convergent improper integral. See "improper integral" in the index of your calculus textbook.
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As you wrote $$\int_0^a \log(x)\,dx=a (\log (a)-1)$$ Now, suppose that you want that $$a (\log (a)-1)=b$$ Rewrite first $$a (\log (a)-1)=e\, \frac a e \,\log \left(\frac{a}{e}\right) $$ Let $t=\frac{a}{e}$ to make $$t\log(t)=\frac be$$ Exponentiate $$t^t=e^{b/e}=k$$ and look at the definition of Lambert function.
So $$t^t=k \quad \implies \quad t=W(k)=e^{W\left(\frac{b}{e}\right)}\quad \implies \quad a=e\,e^{W\left(\frac{b}{e}\right)}=e^{1+W\left(\frac{b}{e}\right)}$$
Would you agree that the following is true: $$\int_{x=0}^\infty e^{-x} \, dx = 1. \tag{1}$$
After all, we know that the antiderivative of $e^{-x}$ is $-e^{-x}$, hence for any positive real $N$, we have $$\int_{x=0}^N e^{-x} \, dx = \Bigl[-e^{-x}\Bigr]_{x=0}^N = -e^{-N} + e^{-0} = 1 - e^{-N}. \tag{2}$$ Consequently, the improper integral $(1)$ has value $$\lim_{N \to \infty} 1 - e^{-N} = 1 - 0 = 1. \tag{3}.$$
If you view this geometrically, it means that the area under the curve $y = e^{-x}$ on the interval $x \in [0,\infty)$ is $1$. So by reflection symmetry, it also means that the area under the curve $y = e^x$ on the interval $x \in (-\infty, 0]$ is also $1$. And then by a further reflection about the line $y = x$, this implies that the absolute area under the curve of $y = \log x$ on the interval $x \in (0, 1]$ is also $1$: see the diagram below.
The blue curve is $y = e^x$. The orange curve is the inverse function, $y = \log x$. The blue shaded region is has the area $$\int_{x=-\infty}^0 e^x \, dx = 1$$ as demonstrated above. The orange shaded region, being the reflection of the blue region, also has absolute area $1$; however, as it lies below the $x$-axis, its signed area is represented by the integral $$\int_{x=0}^1 \log x \, dx = -1. \tag{4}$$
This in turn implies that the value of $a$ for which $$\int_{x=0}^a \log x \, dx = 1 \tag{5}$$ is the value for which $$\int_{x=1}^a \log x \, dx = 2, \tag{6}$$ since $\log 1 = 0$. It is not difficult to see that such a value does in fact exist. We would use integration by parts with the choice $$u = \log x, \quad du = \frac{1}{x} \, dx, \\ dv = dx, \quad v = x, \tag{7}$$ to compute the indefinite integral $$\int \log x \, dx = uv - \int v \, du = x \log x - \int x \cdot \frac{1}{x} \, dx = x \log x - x + C.$$ Hence $$\int_{x=1}^a \log x \, dx = \Bigl[ x \log x - x \Bigr]_{x=1}^a = a \log a - a - (1 \log 1 - 1) = a \log a - a + 1. \tag{8}$$ As we require this to be equal to $2$, we have the equation $$a (-1 + \log a) = 1, \tag{9}$$ which has no elementary closed form solution but has the approximate numerical solution $a \approx 3.59112$.