How do you solve the equation #log_2n=1/4log_2 16+1/2log_2 49#?

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Answer 1 · 2016-10-30T02:58:49

#n=14#

#log_2 n =1/4 log_2 16 + 1/2 log_2 49#

First use the log rule #alogx=logx^a#.

#log_2 n =log_2 16^(1/4)+ log_2 49^(1/2)#

#16^(1/4)=root(4)16=2# and #49^(1/2)=sqrt49=7#

#log_2 n = log_2 2 +log_2 7#

Use the log rule #logx +logy = logxy# to condense the log.

#log_2 n = log_2 (2 *7)#

#log_2 n = log_2 14#

#n=14#