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How do you solve #log_7n=2/3log_7 8#?

1 Answer

Nov 24, 2016

#n =4#

Explanation:

#log_7 n = 2/3log_7 8" "larr# use the log power law

#log_7 n = log_7 8^(2/3)#

#:. n = 8^(2/3)" "larr logA = logB hArr A = B#

#n = (root3 8)^2#

#n = 2^2#

#n = 4#

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