The graph of #y = log_("1/a")x# is the same as the graph of? 1. #y = -a^x# 2. #y = log_ax# 3. #y = (1/a)^x# 4. #y = -log_ax# 5. #y = log_a(-x)#

Question

Answer is 4. #y = -log_ax#.

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Answer 1 · 2018-05-10T06:45:49

#4.y=-log_ax#

As #y=log_(1/a)x#, we have #(1/a)^y=x#

or #1/a^y=x#

or #a^y=1/x#

or #log_a(1/x)=y#

or #log_a1-log_ax=y#, but #log_a1=0#

Hence #-log_a x=y# i.e. answer is 4