find-the-solution-of-log-2-x-2-x-3-lt-2-

Question Number 87194 by john santu last updated on 03/Apr/20

find the solution of

\frac{∣ \log_{2} (x) + 2 ∣}{x - 3} < 2

Commented by TANMAY PANACEA. last updated on 03/Apr/20

i s i t (2 + {l o g}_{2} x) o r {l o g}_{2} (x + 2)

Commented by john santu last updated on 03/Apr/20

2 + \log_{2} (x) sir

Commented by john santu last updated on 03/Apr/20

i {can}^{^{'}} t answer that

Commented by john santu last updated on 03/Apr/20

the choice answer is

a) x < 2^{3} or x > 2^{8}

b) 3 < x < 8

c) x < 3 or x > 8

d) 2^{3} < x < 2^{8}

e) 0 < x < 2^{3} or x > 2^{8}

Commented by jagoll last updated on 03/Apr/20

may be the question is

∣ \log_{2} (\frac{x + 2}{x - 3}) ∣ < 2

Commented by jagoll last updated on 03/Apr/20

- 2 < \log_{2} (\frac{x + 2}{x - 3}) < 2

2^{- 2} < \frac{x + 2}{x - 3} < 2^{2}

\Rightarrow {\begin{cases} \frac{x + 2}{x - 3} > \frac{1}{4} \\ \frac{x + 2}{x - 3} < 4 \end{cases}

Answered by TANMAY PANACEA. last updated on 03/Apr/20

\frac{∣ 2 + {l o g}_{2} x ∣}{x - 3} - 2 < 0

1) x \neq 3 a n d x \neq 0

a s s u m e x = 2^{k} k > 0

\frac{∣ 2 + k ∣}{2^{k} - 3} - 2

= \frac{2 + k}{2^{k} - 3} - 2

w h e n k = 1

\frac{2 + 1}{2 - 3} - 2 < 0

w h e n k = 2

\frac{2 + 2}{4 - 3} - 2 > 0 (n o t e h e r e)

w h e n k = 3

\frac{2 + 3}{8 - 3} - 2 < 0

k = 4

\frac{2 + 4}{16 - 3} - 2 < 0

Answered by MJS last updated on 03/Apr/20

\frac{∣ 2 + \log_{2} x ∣}{x - 3} < 2 \Rightarrow x \in R ∖ {0, 3}

(1) 2 + \log_{2} x ⩾ 0 \Rightarrow x ⩾ \frac{1}{4}

\frac{2 + \log_{2} x}{x - 3} < 2

(1.1) \frac{1}{4} ⩽ x < 3 \Rightarrow x - 3 < 0

256 x > 4^{x} \Rightarrow \approx . 00392758 < x <\approx 5.18752

\Rightarrow \frac{1}{4} ⩽ x < 3 ∙

(1.2) x > 3 \Rightarrow x - 3 > 0

256 x < 4^{x} \Rightarrow x <\approx . 00392758 \lor x >\approx 5.28752

\Rightarrow x >\approx 5.28752 ∙

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

\frac{- (2 + \log_{2} x)}{x - 3} < 2 always true because lhs < 0

\Rightarrow x < 0 \lor 0 < x < \frac{1}{4} ∙

\Rightarrow x < 0 \lor 0 < x < 3 \lor x >\approx 5.28752

Commented by MJS last updated on 03/Apr/20

even if the question is wrong, we can solve it

Commented by jagoll last updated on 03/Apr/20

but x < 0 not valid sir

\log_{2} (x) \Rightarrow must be x > 0

Commented by MJS last updated on 03/Apr/20

you are right \dots but ∣ 2 + \log_{2} x ∣ \in R \forall x \neq 0 \dots

I^{'} ll look into this again later

Commented by MJS last updated on 03/Apr/20

\log_{2} x = \frac{\ln x}{\ln 2}

for x \in C

\frac{\ln x}{\ln 2} = \frac{\ln (r e^{i θ})}{\ln 2} = \frac{\ln r + i θ}{\ln 2}

∣ \frac{\ln r + i θ}{\ln 2} ∣= \frac{\sqrt{θ^{2} + {(\ln r)}^{2}}}{\ln 2}

for x \in R^{-} : θ = π \Rightarrow ∣ 2 + \log_{2} x ∣⩾ 0

\Rightarrow in our case for x < 0 the equation is true

Commented by john santu last updated on 03/Apr/20

ooo if x \in C sir .

so the choice answer nothing correct

sir