The points $P$ and $Q$ lie on the curve with equation $y=\mathrm{e}^{\frac{1}{2} x}$. The $x$-coordinates of $P$ and $Q$ are $\ln 4$ and $\ln 16$ respectively. a Find an equation for the line $P Q$. b Show that this line passes through the origin $O$. c Calculate the length, to 3 significant figures, of the line segment $P Q$.

Step 1 ：Let P = $(\ln 4, \mathrm{e}^{\frac{1}{2} \ln 4})$ and Q = $(\ln 16, \mathrm{e}^{\frac{1}{2} \ln 16})$. Then P = $(\ln 4, 2)$ and Q = $(\ln 16, 4)$.

Step 2 ：Using the two-point form of a line, we have $y - 2 = \frac{4 - 2}{\ln 16 - \ln 4}(x - \ln 4)$.

Step 3 ：Simplifying, we get $y - 2 = \frac{2}{\ln 16 - \ln 4}(x - \ln 4)$.

Step 4 ：Using the properties of logarithms, we have $\ln 16 - \ln 4 = \ln \frac{16}{4} = \ln 4$.

Step 5 ：Substituting, we get $y - 2 = \frac{2}{\ln 4}(x - \ln 4)$.

Step 6 ：Thus, the equation of the line PQ is $\boxed{y - 2 = \frac{2}{\ln 4}(x - \ln 4)}$.

Step 7 ：To show that the line passes through the origin, we substitute $x = 0$ and $y = 0$ into the equation.

Step 8 ：Substituting, we get $0 - 2 = \frac{2}{\ln 4}(0 - \ln 4)$.

Step 9 ：Simplifying, we get $-2 = -2$, which is true. Thus, the line passes through the origin.

Step 10 ：To find the length of PQ, we use the distance formula: $\sqrt{(\ln 16 - \ln 4)^2 + (4 - 2)^2}$.

Step 11 ：Simplifying, we get $\sqrt{(\ln 4)^2 + 2^2}$.

Step 12 ：Calculating, we get $\sqrt{16} = 4$.

Step 13 ：Thus, the length of PQ is $\boxed{4}$ to 3 significant figures.