log4 128-log4 8

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log4 128-log4 8

Guest Apr 10, 2015

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Best Answer

In general log_na - log_nb = log_n(a/b) so log₄(128) - log₄(8) = log₄(128/8) = log₄(16)

Also, when n^a = b then a = log_nb so log₄(16) = 2 (because 4² = 16)

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Alan Apr 10, 2015

Alan · Answer 1 · 2015-04-10T08:06:49

In general log_na - log_nb = log_n(a/b) so log₄(128) - log₄(8) = log₄(128/8) = log₄(16)

Also, when n^a = b then a = log_nb so log₄(16) = 2 (because 4² = 16)

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