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

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Answer 1 · 2016-11-24T10:00:38

$n =4$

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

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

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

$n = (root3 8)^2$

$n = 2^2$

$n = 4$

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