Compare $\ln(\pi)$ and $\pi-2$ without calculator Unlike the famous question of comparing $e^\pi$ and $\pi^e$, which I solved almost instantly, I am stuck with this problem.  My thought was the following.
Since exponential function is order-preserving, we exponentiate both terms and get $\pi$ and $e^{\pi-2}$.  Then we study the function $f(x) = e^{x-2} - x$ or the function $g(x) = \frac{e^{x-2}}{x}$, and compare them with zero and one respectively.  I tried both.  But both involve solving the equation
$$e^{x-2} = x.$$
I tried Lagrange error terms and have
$$f(x) = -1 + \frac{(x-2)^2}{2!} + R_2(x-2),$$
where
$$\frac{(x-2)^3}{3!} \le R_2(x-2) \le \frac{e^{x-2}}{3!} (x-2)^3.$$
It is easy to see that the equation have a root between $3$ and $2 + \sqrt2$.  But I don't know how close it is to $\pi$.  It is to provide some lower bounds since we can plug in some values and calculate to show that $f(x) > 0$ for such values.  But for the upper bound, it is hard to calculate by hands since it has the $e^{x-2}$ factor.  At my best attempt by hand, I showed that $f(3.15) > 0$.  All it entails is that for all $x \ge 3.15$, $e^{x-2}$ is greater than $x$.  But it tells nothing about the other side.
Then I looked at the calculator and find that $e^{\pi-2} < \pi$.
I also tried Newton-Raphson iteration, but it involves a lot of exponentiation which is hard to calculate by hand and also involves approximation by themselves.  And I don't know how fast and close the iteration converges to the true root of the equation.
Any other hint for comparing these two number purely by hand?

By hand I took $e^{\pi -2}<e\cdot e^{0.1416}<(2.7183)e^{0.1416}.$ Using $B_1=0.1416=0.1+0.04(1+0.04)$ for manual calculation, I computed, to $5$ decimal places, an upper bound $B_2$   for $(B_1)^2/2$ and an upper bound $B_3$ for $B_1B_2/3 $ and an upper bound $B_4$ for $B_1B_3/4,$ etc., until I was sure that the sum of the remaining terms was less than $0.00005,$ to obtain an upper bound $B$ to $4$ decimal places for $e^{0.1416}.$ Then I multiplied $B\times 2.7183$ and got less than $\pi.$
