With a resistor in the emitter, the input resistance looking into the base of the transistor would be (beta + 1)Re. That would reduce the amount of base current. You can't ignore that resistor or the beta of the transistor in calculating the base current in this particular problem.

hgmjr

 

Maybe there was a Ve I missed. If I had that, that would need to be subtracted also.

 

Absolutely.

hgmjr

 

Thanks for helping me out.

 

That is what we are here to try to do.

hgmjr