Question

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

Answer 1

`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`