Logarithm Problem : Find the number of real solutions of the equation 2 log 2 </msu

skottyrottenmf

skottyrottenmf

Answered question

2022-05-19

Logarithm Problem : Find the number of real solutions of the equation
2 log 2 log 2 x + log 1 2 log 2 ( 2 2 x ) = 1
My approach :
Solution : Here right hand side is constant term so convert it into log of same base as L.H.S. therefore, 1 can be written as log 2 log 2 4
2 log 2 log 2 x + log 1 2 log 2 ( 2 2 x ) = log 2 log 2 4 log 2 log 2 x 2 log 2 log 2 ( 2 2 x ) = log 2 log 2 4
log 2 log 2 x 2 log 2 ( 2 2 x ) = log 2 log 2 4
log 2 x 2 log 2 ( 2 2 x ) = log 2 4
Please suggest whether is it the right approach... thanks...

Answer & Explanation

soymmernenx

soymmernenx

Beginner2022-05-20Added 10 answers

I think you went off the rails in the second line. Here's what I get
2 log 2 log 2 x = log 2 log 2 2 x
so that
2 log 2 log 2 x + log 1 / 2 log 2 ( 2 2 x ) = log 2 log 2 2 x log 2 ( 2 2 x )
and the equation becomes
log 2 log 2 2 x log 2 ( 2 2 x ) = 1 log 2 2 x log 2 ( 2 2 x ) = 2
or
log 2 2 x = 2 log 2 2 3 / 2 + 2 log 2 x log 2 2 x 2 log 2 x 3 = 0
This implies that log 2 x = 3 or log 2 x = 1. In the former case, we have that x = 8 in the latter, we have x = 1 / 2 . However, in the latter case, we have a false solution, as log 2 log 2 ( 1 / 2 ) = log 2 ( 1 ) which is outside the realm of the reals. Thus, the only solution is at x = 8

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