Gary, in your Joule Thief circuit with the added LDR transistor, R2 turns on the transistor and R1 turns it off.
If the battery is 2.7V, R1 is a high value and R2 is very low at 5k ohms then the base current in Q1 is (2.7V - 0.7V)/5k= 0.4mA. Then the datasheet of the transistor says it will saturate very well with an output current as high as 4mA which is much more current that a Joule Thief needs.
Maybe your transistor has its base destroyed if the 100k pot resistance was turned too low or maybe the transistor emitter and collector are wrongly swapped.
If the battery is 2.7V, R1 is a high value and R2 is very low at 5k ohms then the base current in Q1 is (2.7V - 0.7V)/5k= 0.4mA. Then the datasheet of the transistor says it will saturate very well with an output current as high as 4mA which is much more current that a Joule Thief needs.
Maybe your transistor has its base destroyed if the 100k pot resistance was turned too low or maybe the transistor emitter and collector are wrongly swapped.