Along an exchanger, the gap between the two streams changes from one end to the other. The correct average of that changing gap is not the arithmetic mean but the log mean: $\Delta T_{lm} = \dfrac{\Delta T_1 - \Delta T_2}{\ln(\Delta T_1 / \Delta T_2)}$, read delta-T log-mean equals delta-T-one minus delta-T-two, over the natural log of delta-T-one over delta-T-two.

The subscripts number the ENDS of the shell, not the streams: $\Delta T_1$ is the temperature difference at one end and $\Delta T_2$ the difference at the other, both in kelvin. Which temperatures pair up is set entirely by the plumbing, and this is the step that is actually examined.

In counterflow, the streams enter at opposite ends, so the hot INLET faces the cold OUTLET: $\Delta T_1 = T_{h,in} - T_{c,out}$ and $\Delta T_2 = T_{h,out} - T_{c,in}$. In parallel flow both enter together, so the two inlets are one end and the two outlets the other: $\Delta T_1 = T_{h,in} - T_{c,in}$, $\Delta T_2 = T_{h,out} - T_{c,out}$. Here h is the hot stream, c the cold, and in and out mean entering and leaving.

Two nuggets worth keeping. First, the ratio inside the logarithm is kelvin over kelvin — dimensionless, and it has to be, because no logarithm will accept a unit. Second, run the identical streams both ways and counterflow always returns the larger log mean, every time. That is not a coincidence of the arithmetic; it is why counterflow is what gets built. And when the two terminal differences happen to be equal, the formula quietly hands back that difference itself — no logarithm required.