find-solution-set-of-inequality-log-2-x-2-3x-1-lt-log-2-x-2-3-x-

Question Number 100666 by bobhans last updated on 28/Jun/20

find solution set of inequality

{(\log_{2} x - 2)}^{3 x - 1} < {(\log_{2} x - 2)}^{3 - x}

Commented by Rasheed.Sindhi last updated on 28/Jun/20

{(\log_{2} x - 2)}^{3 x - 1} < {(\log_{2} x - 2)}^{3 - x}

\log_{2} x - 2 \neq 0, 1 \Rightarrow 3 x - 1 < 3 - x

\Rightarrow 4 x < 4 \Rightarrow x < 1

Commented by bramlex last updated on 28/Jun/20

i t h i n k n o t c o r r e c t s i r

Commented by Rasheed.Sindhi last updated on 28/Jun/20

{Y o u}^{'} r e r i g h t s i r!

Answered by bramlex last updated on 28/Jun/20

\Leftrightarrow (\log_{2} x - 2 - 1) (3 x - 1 - (3 - x)) < 0

(\log_{2} x - 3) (4 x - 4) < 0

4 (\log_{2} x - 3) (x - 1) < 0

c a s e 1 \Rightarrow \log_{2} x - 3 < 0 \land x - 1 > 0

x < 8 \land x > 1 \Rightarrow 1 < x < 8

c a s e 2 \Rightarrow \log_{2} x - 3 > 0 \land x - 1 < 0

x > 8 \land x < 1 \Rightarrow x = \emptyset

s o l u t i o n (1) \cup (2) \Rightarrow 1 < x < 8

Commented by bemath last updated on 28/Jun/20

sir bramlex it should be (\log_{2} x - 2 - 1)

n o t (\log_{2} x - 2 + 1)

Commented by bramlex last updated on 28/Jun/20

o o y e s . . y o u r a r e r i g h t

Answered by 1549442205 last updated on 28/Jun/20

the condition for the given inequality

is defined as {\begin{cases} x > 0 \\ \log_{2} x - 2 > 0 \end{cases} \Leftrightarrow x > 4 . Then

The given inequality is equivalent to

[\log_{2} x - 2 - 1) [(3 x - 1) - (3 - x)] < 0

\Leftrightarrow (\log_{2} x - 3) (4 x - 4) < 0 \Leftrightarrow (\log_{2} x - 3) (x - 1) < 0

\Leftrightarrow [_{{\begin{cases} \log_{2} x - 3 < 0 \\ x - 1 > 0 \end{cases} \Leftrightarrow {\begin{cases} 4 < x < 8 \\ x > 1 \end{cases} \Leftrightarrow 4 < x < 8}^{{\begin{cases} \log_{2} x - 3 > 0 \\ x - 1 < 0 \end{cases} \Leftrightarrow {\begin{cases} x > 8 \\ x < 1 \end{cases} \Rightarrow has no solutions}

Thus, solution set of the inequality is

the interval (4, 8)