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1 Answer
NickTheTurtle
Apr 7, 2017

#-oo< x<1#

Explanation:

Solve the function for #y#: #y=log_a(a-a^x)#.

Since the logarithm can only accept values greater than #0#, #a-a^x>0#, or #a^x< a#. We can take the natural logarithm of both sides to get #xln(a)< ln(a)#.

Since #a>1#, #ln(a)>0#. We can divide both sides by #ln(a)# without worrying about flipping the inequality sign or dividing by #0#. Therefore, #xln(a)< ln(a)# becomes #x<1#.

Thus, the domain is #-oo< x<1#.

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