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Answer 1 · 2015-08-21T13:26:28

#x = 28/3#

#-log3(x-9)=0#
#log3(x-9)=0#
#Rightarrow 3(x-9) = 1#(since #log 1 = 0# regardless of base)
#x-9 = 1/3#
#x = 28/3#

Answer 2 · 2015-08-21T21:56:30

If the intended question is: Solve #-log_3(x-9)=0#, then the solution is #x=10#

#-log_3(x-9)=0# if and only if #log_3(x-9)=0#

Change to exponential form: #(x-9) = 3^0#.

So #x-9 = 1#, and #x = 10#.