How do you solve $log_2 (x - 2) + 5 = 8 - log_2 4$ ?

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Answer 1 · 2015-07-18T10:59:41

x=4

$log(ab)=loga+logb$ ____(1)

$log_a(b)=x =>a^x=b$ _______(2)

$log_2(x-2)+5=8-log_2(4)$

$=>log_2(x-2)+log_2(4)=3$

$=>log_2((x-2)4)=3$ [since from (1)]

$=>2^3=4(x-2)$ [since from (2)]

$=>8=4x-8$

$=>4x=16$

$=>x=4$

How do you solve log_2 (x - 2) + 5 = 8 - log_2 4 log_2 (x - 2) + 5 = 8 - log_2 4?