How do you solve #2^(2n)<=1/16#?

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Answer 1 · 2016-10-13T21:06:59

#n <=-2#

#log(2^(2n)) <= log(1/16)#

#2nlog2 <= log(2^(-4))#

#2nlog2 <=log(2^(-4))#

#2nlog2 <=-4log2#

#2nlog2 + 4log2 <= 0#

#log2(2n + 4) <= 0#

#2n + 4 <= 0#

#n <= -2#

Hopefully this helps!

Answer 2 · 2016-10-13T21:08:46

#n<=-2#

#2^(2n)<=1/16#

#log(2^(2n))<=log(1/16)#

Using the power rule #-# (#log_b(x^y) = ylog_b(x)#) #-# we can rewrite this inequality as:

#2nlog(2)<=log(1/16)#

#2n<=log(1/16)/(log(2))#

#2n<=-4#

#n<=-4/2#

#n<=-2#