How do you solve #Log_[2]x + log_[2]x + 1= 2 - log_[2] (1/3)#?

1 Answer
Vikki
Dec 7, 2015

#x= sqrt 6#

Explanation:

Given #log_2x + log_2x+ 1= 2-log_2(1/3)#

Step 1: Manipulate the equation to have all logarithm on one side, like so

#log_2x + log_2x +log_2(1/3) = 2-1#

Step 2: Use the sum to product logarithmic properties
#logAB= log A + log B#

#log_2x + log_2x +log_2(1/3) = 1#
#log_2(x*x*1/3)=1#
#log_2 (1/3 x^2) = 1#
Change logarithmic equation to exponential form since
#log_ax = y hArr a^y = x#
#2^1 = 1/3 x^2#
#6= x^2#
#x= +-sqrt6#

Only positive answer would work because the domain #logx# exist if #x>0#

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