Your options 2. and 3. are not valid, because the voltage across the OFF diode is not negative.
For your ideal diode to be OFF it must have a negative voltage and zero current, while it should have a positive current and zero voltage when it is ON. Always check for these 2 conditions.
You are stating that your ideal diode has zero current and zero voltage, which is none of its 2 available states when you are analyzing your circuit.
(Please note that this is being theoretically strict (as we should be when analyzing a circuit of ideal components). Practically your diode can have V = 0, I = 0, but it is not to much use when you are analyzing your circuit)
Here's the way I'd analyze this circuit:
1) By inspection, the emitter resistor is smaller than the collector resistor. So short C-E and calculate the voltage at that junction. I get about 4.13V.
2) The Base voltage is set at 6V. The voltage from above (4.13V) means that the E-B junction is forward biased. That changes things. Note that you need to do the above calculation to determine if, in fact, the junction is forward-biased.
3) The Emitter voltage is the Base voltage minus Vbe. That is: 6V - (0.7) = 5.3V. The current through the Emitter resistor is that voltage divided by the resistance: 5.3 / 3300 = ~1.61 mA.
4) Assume that the transistor is saturated. We'll do the calculations based on that assumption, then go back and check the assumption once we have some numbers to work with.
5) The collector voltage is about Vemitter plus the saturation voltage. I would normally assume saturation voltage at about 150 mV but your text says 200 mV, so use that. Ve + 0.2 = 5.5V.
6) Calculate the collector current. (10 - 5.5) / 4700 = ~0.957 mA
7) The Base current is the Emitter current minus the Collector current. 1.61 - 0.957 = ~0.649 mA
Now check to see if the transistor is in saturation. Most small-signal transistors have a Hfe of anywhere from 40 to 200. Let's use the worst-case value of 40.
If the transistor was NOT saturated, the collector current would be greater than 0.649 mA * 40 = ~25.9 mA. But we already know the collector current is about 0.957 mA. Therefore, the transistor is saturated and the above calculations hold.
Note that it took far longer to type this out than it did to calculate.
Best Answer
When analyzing complex circuits, you need to be able to break down the overall circuit into sub-circuits which you have already done. Then you need to understand how each sub-circuit functions. You can do this by simulation or do some research and find similar circuits. The next step is to find some equations if possible to describe the sub-circuits.
Superposition is your friend, remove and add different parts of the circuit or substitute voltages and currents. Observe how the summing circuit would function without Q2. Then add it in and see what happens. Also simulate the "green" log circuit with a sin wave input. Run frequency sweeps or an AC analysis if the design concerned with the frequency domain. This circuit has a fixed voltage which makes it nice because half of the circuit is running at a DC fixed value, that makes an equation analysis easier.
Here is some info I've found on log circuits: Maxim integrated Log IC