First note that #(2/3)^x-9# is well defined for any Real value of #x#. So the domain is the whole of #RR#, i.e. #(-oo, oo)#
Since #0 < 2/3 < 1#, the function #(2/3)^x# is an exponentially decreasing function which takes large positive values when #x# is large and negative, and is asymptotic to #0# for large positive values of #x#.
In limit notation, we can write:
#lim_(x->-oo) (2/3)^x = -oo#
#lim_(x->oo) (2/3)^x = 0#
#(2/3)^x# is continuous and strictly monotonically decreasing, so its range is #(0, oo)#.
Subtract #9# to find that the range of #(2/3)^x# is #(-9, oo)#.
Let:
#y = (2/3)^x-9#
Then:
#y+9 = (2/3)^x#
If #y > -9# then we can take logs of both sides to find:
#log(y+9) = log((2/3)^x) = x log(2/3)#
and hence:
#x = log(y+9)/log(2/3)#
So for any #y in (-9, oo)# we can find a corresponding #x# such that:
#(2/3)^x-9 = y#
That confirms that the range is the whole of #(-9, oo)#.
