How do you solve $log_2(4y-10)>=log_2(y-1)$ ?

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Answer 1 · 2016-10-18T17:06:39

$log_2(4y - 10) - log_2(y - 1) ≥ 0$

$log_2((4y - 10)/(y - 1)) ≥ 0$

$(4y - 10)/(y - 1) ≥ 2^0$

$4y - 10 ≥ 1(y - 1)$

$3y ≥ 9$

$y ≥ 3$

Hopefully this helps!

How do you solve log_2(4y-10)>=log_2(y-1)log_2(4y-10)>=log_2(y-1)?