Understanding CAS variables

[HuangDangDaniel]

Han, thank you for the info, I tried it with the function app and it still has the same behavior unless I'm using something like solve(e^a=5,a) (returns ln{5}). Anything that isn't the variable variable I'm solving for still evaluates to 0, so ((a*b)=5,a) will still evaluate to [], because b is treated as 0. Would it be best to just wipe to software and start over at this point? Or is there something I'm doing wrong or something else to try?