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How do you evaluate #log_343 (1/7)#?

1 Answer

Nov 13, 2016

#-1/3#

Explanation:

Let #n = log_343 (1/7)#

#343^n=1/7#

Rewritten using all bases of 7:
#(7^3)^n=7^(-1)#

#7^(3n)=7^(-1)#

#3n=-1#

#n=-1/3#

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