How do you evaluate #log_7 (1/49)#?

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Answer 1 · 2018-05-19T22:30:44

#log_7(1/49)=x#

#7^x=(1/49)#

Answer 2 · 2018-05-19T22:46:51

Ans -2

#log_7 (1/49)# = y

#log_7 (1/(7^2))# =y

#log_7 (7^-2)# =y

y = -2