How do you solve #log_4 7 + 2 log_4 x = log_4 2#?

1 Answer
Alan P.
Dec 14, 2015

#x=sqrt(2/7)#

Explanation:

Given:
#color(white)("XXX")log_4(7)+2log_4(x)=log_4(2)#

Remember
#color(white)("XXX")#log multiplication rule: #log_b(p*q) =log_b(p)+log_b(q)#
#color(white)("XXX")#log power rule: #log_b(s^t) = t*log_b(s)#

Therefore the given equation can be rewritten as
#color(white)("XXX")log_4(7x^2)=log_4(2)#

from which it follows that
#color(white)("XXX")7x^2=2#

#color(white)("XXX")x^2=2/7#

#color(white)("XXX")x=sqrt(2/7)#
#color(white)("XXXXXX")#we can ignore the negative root as extraneous since #sqrt(x)# requires #x>=0# for Real solutions

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