How-many-positive-integers-x-satisfy-log-x-8-x-2-4-lt-7-log-2-8-x-

Question Number 118511 by ZiYangLee last updated on 18/Oct/20

How many positive integers x satisfy

\log_{\frac{x}{8}} \frac{x^{2}}{4} < 7 + \log_{2} \frac{8}{x}

Answered by mr W last updated on 18/Oct/20

2 \times \frac{\ln \frac{x}{2}}{\ln \frac{x}{8}} < 7 + \frac{\ln \frac{8}{x}}{\ln 2}

2 \times \frac{\ln \frac{x}{2}}{\ln \frac{x}{2} - 2 \ln 2} < 7 + \frac{2 \ln 2 - \ln \frac{x}{2}}{\ln 2}

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

\frac{2 t}{t - 2 \ln 2} < \frac{9 \ln 2 - t}{\ln 2}

\Rightarrow t^{2} - (9 \ln 2) t + 18 {(\ln 2)}^{2} < 0

\Rightarrow t = \frac{9 \ln 2 \pm 3 \ln 2}{2} = 3 \ln 2, 6 \ln 2

\Rightarrow 3 \ln 2 < t < 6 \ln 2

\Rightarrow \ln 2^{3} < \ln \frac{x}{2} < \ln 2^{6}

\Rightarrow 2^{4} = 16 < x < 2^{7} = 128

\Rightarrow 17 ⩽ x ⩽ 127 \Rightarrow 111 s o l u t i o n s

Answered by Lordose last updated on 18/Oct/20

\log_{\frac{x}{8}} {(\frac{x}{2})}^{2} < 7 + \log_{2} \frac{8}{x}

changing the base of logarithm,

\frac{\log_{2} {(\frac{x}{2})}^{2}}{\log_{2} (\frac{x}{2} \times \frac{1}{4})} < 7 + \log_{2} (\frac{2}{x}) + \log_{2} 2^{2}

\frac{{2 \log}_{2} (\frac{x}{2})}{\log_{2} (\frac{x}{2}) - 2} < 9 - \log_{2} (\frac{x}{2})

set \log_{2} (\frac{x}{2}) = y

\frac{2 y}{y - 2} < 9 - y

2 y < (9 - y) (y - 2)

2 y < 9 y - 18 - y^{2} + 2 y

0 < - (y^{2} - 9 y + 18)

0 > y^{2} - 9 y + 18

0 > (y - 3) (y - 6)

(y - 3) (y - 6) < 0

y < 3 and y < 6

x < 16 and x < 128

Answered by 1549442205PVT last updated on 18/Oct/20

We need the condition x > 0, x \neq 8

\log_{\frac{x}{8}} \frac{x^{2}}{4} < 7 + \log_{2} \frac{8}{x} \Leftrightarrow \log_{\frac{x}{8}} [{(\frac{x}{8})}^{2} . 16]

< 7 + \log_{2} 8 - \log_{2} x \Leftrightarrow \log_{\frac{x}{8}} {(\frac{x}{8})}^{2} + \log_{\frac{x}{8}} 16

< 7 + 3 - \log_{2} x \Leftrightarrow 2 + {4 \log}_{\frac{x}{8}} 2 < 10 - \log_{2} x

\Leftrightarrow \frac{4}{\log_{2} \frac{x}{8}} < 8 - \log_{2} x \Leftrightarrow \frac{4}{\log_{2} x - 3} < 8 - \log_{2} x

\Leftrightarrow \frac{4}{\log_{2} x - 3} + \log_{2} x - 8 < 0 . Put \log_{2} x = t

\Leftrightarrow \frac{4}{t - 3} + t - 8 < 0 \Leftrightarrow \frac{t^{2} - 11 t + 28}{t - 3} < 0

\Leftrightarrow \frac{(t - 4) (t - 7)}{t - 3} < 0

\Leftrightarrow t \in (- \infty, 3) \cup (4, 7)

i) \log_{2} x < 3 \Leftrightarrow 0 < x < 8

ii) 4 < \log_{2} x < 7 \Leftrightarrow 2^{4} < x < 2^{7} \Leftrightarrow 16 < x < 128

Thus, the roots of given inequality are :

x \in (0, 8) \cup (16, 128)

\boldsymbol Since \boldsymbol x \in \boldsymbol N^{*}, \boldsymbol x \in {1, 2, \dots, 7} \cup {17, \dots, 127}

\boldsymbol hence \boldsymbol all \boldsymbol there \boldsymbol are 11 \boldsymbol positive \boldsymbol integers

\boldsymbol satisfying \boldsymbol the \boldsymbol given \boldsymbol inequality

Commented by mr W last updated on 18/Oct/20

v e r y c o r r e c t!

v e r y c o r r e c t!

Commented by 1549442205PVT last updated on 18/Oct/20

Thank Sir . You are welcome

Thank Sir . You are welcome

Commented by ZiYangLee last updated on 19/Oct/20

PERFECT! ★ ★

PERFECT! ★ ★