How do you expand $log_2 (16/sqrt(x-1))$ ?

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Answer 1 · 2016-06-28T14:53:29

$=log_2(16) - log_2(sqrt(x - 1))$

$=4 - log_2(x - 1)^(1/2)$

$=4 - 1/2log_2(x - 1)$

Hopefully this helps!

How do you expand log_2 (16/sqrt(x-1))log_2 (16/sqrt(x-1))?