Solving the equation $10^{x-1} = x$ exactly is difficult.
But you don't have to solve it exactly in order to figure out how many times the two graphs meet.
First, note that $y=\log_{10}x$ is only defined on the positive real numbers. So we can restrict ourselves to $(0,\infty)$.
Then, consider the function $f(x) = x-1-\log_{10}x$. The derivative of the function is
$$f'(x) = 1 - \frac{1}{\ln(10)x}.$$
The derivative is positive if $x\gt \frac{1}{\ln(10)}$, and negative if $x\lt \frac{1}{\ln(10)}$. That means that the function $f(x)$ is decreasing on $(0,\frac{1}{\ln(10)})$, and is increasing on $(\frac{1}{\ln 10},\infty)$.
As $x\to 0^+$, we have $f(x)\to\infty$ (since $\log_{10}(x)\to-\infty$). At $x=\frac{1}{\ln(10)}$, we have $f(x)\approx -0.2035$; and as $x\to\infty$, $f(x)\to\infty$. So the function crosses the $x$-axis somewhere between $0$ and $\frac{1}{\ln(10)}\approx 0.4343$, and then again somwhere after $\frac{1}{\ln(10)}$ (well, at $x=1$, to be precise). And that's it.
So there are exactly two intersections.