Question

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

Answer 1

`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`