How many positive integers x satisfy log_(x/8) (x^2 /4)<7+log


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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
	$We need the condition x>0,x≠8 log_(x/8) (x^2 /4)<7+log_2 (8/x)⇔log_(x/8) [((x/8))^2 .16] <7+log_2 8−log_2 x⇔log_(x/8) ((x/8))^2 +log_(x/8) 16 <7+3−log_2 x⇔2+4log_(x/8) 2<10−log_2 x ⇔(4/(log_2 (x/8)))<8−log_2 x⇔(4/(log_2 x−3))<8−log_2 x ⇔(4/(log_2 x−3))+log_2 x−8<0.Put log_2 x=t ⇔(4/(t−3))+t−8<0⇔((t^2 −11t+28)/(t−3))<0 ⇔(((t−4)(t−7))/(t−3))<0 ⇔t∈(−∞,3)∪(4,7) i)log_2 x<3⇔0<x<8 ii)4<log_2 x<7⇔2^4 <x<2^7 ⇔16<x<128 Thus,the roots of given inequality are: x∈(0,8)∪(16,128) Since x∈N^∗ ,x∈{1,2,...,7}∪{17,...,127} hence all there are 11 positive integers satisfying the given inequality$
	$We need the condition x > 0, x \neq 8$ $\log_{\frac{x}{8}} \frac{x^{2}}{4} < 7 + \log_{2} \frac{8}{x} \Leftrightarrow \log_{\frac{x}{8}} [{(\frac{x}{8})}^{2} . 16]$ $< 7 + \log_{2} 8 - \log_{2} x \Leftrightarrow \log_{\frac{x}{8}} {(\frac{x}{8})}^{2} + \log_{\frac{x}{8}} 16$ $< 7 + 3 - \log_{2} x \Leftrightarrow 2 + {4 \log}_{\frac{x}{8}} 2 < 10 - \log_{2} x$ $\Leftrightarrow \frac{4}{\log_{2} \frac{x}{8}} < 8 - \log_{2} x \Leftrightarrow \frac{4}{\log_{2} x - 3} < 8 - \log_{2} x$ $\Leftrightarrow \frac{4}{\log_{2} x - 3} + \log_{2} x - 8 < 0 . Put \log_{2} x = t$ $\Leftrightarrow \frac{4}{t - 3} + t - 8 < 0 \Leftrightarrow \frac{t^{2} - 11 t + 28}{t - 3} < 0$ $\Leftrightarrow \frac{(t - 4) (t - 7)}{t - 3} < 0$ $\Leftrightarrow t \in (- \infty, 3) \cup (4, 7)$ $i) \log_{2} x < 3 \Leftrightarrow 0 < x < 8$ $ii) 4 < \log_{2} x < 7 \Leftrightarrow 2^{4} < x < 2^{7} \Leftrightarrow 16 < x < 128$ $Thus, the roots of given inequality are :$ $x \in (0, 8) \cup (16, 128)$ $Since x \in N^{*}, x \in {1, 2, . . ., 7} \cup {17, . . ., 127}$ $hence all there are 11 positive integers$ $satisfying the given inequality$
	Commented bymr W last updated on 18/Oct/20

	$v e r y c o r r e c t!$
	Commented by1549442205PVT last updated on 18/Oct/20

	$Thank Sir . You are welcome$
	Commented byZiYangLee last updated on 19/Oct/20

	$PERFECT! ★ ★$
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