Question

$Ajutor~! \\ Demonstrati: \\ \\ log_{7}14- log_{14}28~apartine \\ lui~( \frac{1}{6} , \frac{1}{12} ) \\ Am ~ajuns~la~:6\ \textless \ \frac{a^2}{a+1} \ \textless \ 12 \\ unde~a=log_{7}2.$

Answer 1

$\log_{\big7}14 - \log_{\big{14}}28 = \log_{\big7}7\cdot 2-\log_{\big{14}}14\cdot 2 = \\ \\ = \log_\big7 7+\log_{\big{7}}2-\Big(\log_{\big{14}}14+\log_{\big{14}}2\Big) = \\ \\ = 1+\log_{\big7}2-\Big(1+\log_{\big{14}}2}\Big) =1+\log_{\big7}2-1-\log_{\big{14}}2}=\\ \\ =\log_{\big7}2-\log_{\big{14}}2} = \log_{\big7}2 - \dfrac{1}{\log_{\big{2}}14}} = \log_{\big7}2 - \dfrac{1}{\log_{\big{2}}7\cdot 2}} = \\ \\ = \log_{\big7}2 - \dfrac{1}{\log_{\big2}7+\log_{\big2}2} = \dfrac{1}{\log_{\big2}7}-\dfrac{1}{\log_{\big2}7+1}$

[tex] \boxed{1}\quad \log_{\big2}4\ \textless \ \log_{\big 2}7\ \textless \ \log_{\big2}8 \Rightarrow 2\ \textless \ \log_{\big2}7\ \textless \ 3\Big|^{-1} \Rightarrow \\ \\ \Rightarrow \dfrac{1}{2} \ \textgreater \ \dfrac{1}{\log_{\big2}7}\ \textgreater \ \dfrac{1}{3} \Rightarrow \dfrac{1}{3} \ \textless \ \dfrac{1}{\log_{\big2}7}\ \textless \ \dfrac{1}{2} \\ \\ \boxed{2}\quad \log_{\big2}4\ \textless \ \log_{\big 2}7\ \textless \ \log_{\big2}8 \Rightarrow 2\ \textless \ \log_{\big2}7\ \textless \ 3 \Big|+1 \Rightarrow \\ \\[/tex]
$\Rightarrow 3\ \textless \ \log_{\big2}7+1\ \textless \ 4 \Big|^{-1} \Rightarrow \dfrac{1}{3} \ \textgreater \ \dfrac{1}{\log_{\big2}7+1}\ \textgreater \ \dfrac{1}{4}\Big|\cdot(-1) \Rightarrow \\ \\ \Rightarrow -\dfrac{1}{3}\ \textless \ -\dfrac{1}{\log_{\big2}7+1}\ \textless \ -\dfrac{1}{4}$

$\\ Adunam \boxed{1} cu \boxed{2}: \\ \\ \Rightarrow \dfrac{1}{3} - \dfrac{1}{3} \ \textless \ \dfrac{1}{\log_{\big2}7}- \dfrac{1}{\log_{\big2}7+1}} \ \textless \ \dfrac{1}{2}-\dfrac{1}{4} \Rightarrow \\ \\ \Rightarrow 0 \ \textless \ \dfrac{1}{\log_{\big2}7}- \dfrac{1}{\log_{\big2}7+1}}\ \textless \ \dfrac{2-1}{4} \Rightarrow 0 \ \textless \ \dfrac{1}{\log_{\big2}7}- \dfrac{1}{\log_{\big2}7+1}}\ \textless \ \dfrac{1}{4} \Leftrightarrow$

$\Leftrightarrow 0\ \textless \ \log_{\big7}14 - \log_{\big{14}}28\ \textless \ \dfrac{1}{4} \Rightarrow \log_{\big7}14 - \log_{\big{14}}28\in \Big(0,\dfrac{1}{4}\Big) \\ \\\bullet \Big(\dfrac{1}{12},\dfrac{1}{6}\Big) \subset\Big(0,\dfrac{1}{4}\Big) \Rightarrow \boxed{\log_{\big7}14 - \log_{\big{14}}28\in \Big(\dfrac{1}{12},\dfrac{1}{6}\Big) }$

Answer 2

[tex]\it log_714 =log_77\cdot2 =log_77+log_72 =1+\dfrac{1}{log_27} \\\;\\ \\\;\\ log_{14}{28} =log_{14}{14\cdot2} =log_{14}14+log_{14}2=1+\dfrac{1}{log_2{14}}= \\\;\\ \\\;\\ =1+\dfrac{1}{log_2{2\cdot7}} =1+\dfrac{1}{1+log_27}[/tex]

Relația din enunț devine:


[tex]\it 1+\dfrac{1}{log_27} -1-\dfrac{1}{1+log_27} = \dfrac{1}{log_27} -\dfrac{1}{1+log_27} = \dfrac{1+log_27 -log_2 7}{log_27(1+log_2 7)} \\\;\\ \\\;\\ = \dfrac{1}{log_27(1+log_2 7)} [/tex]

Expresia aparține intervalului (1/12,  1/6), adică :


$\it \dfrac{1}{12} \ \textless \ \dfrac{1}{log_27(1+log_2 7)} \ \textless \ \dfrac{1}{6} \Rightarrow 6 \ \textless \ log_27(1+log_2 7 \ \textless \ 12$

[tex]\it log_27 \ \textless \ log_28 \Rightarrow log_27\ \textless \ 3 \Rightarrow log_27(1+log_27) \ \textless \ 3\cdot(1+3) =12 \ \ (*) \\\;\\ log_27 \ \textgreater \ log_24 \Rightarrow log_27\ \textgreater \ 2 \Rightarrow log_27(1+log_27) \ \textgreater \ 2\cdot(1+2) =6 \ \ (**) \\\;\\ \\\;\\ (*),\ (**) \Rightarrow 6\ \textless \ log_27(1+log_27) \ \textless \ 12 \Leftrightarrow \dfrac{1}{12} \ \textless \ \dfrac{1}{log_27(1+log_27)}\ \textless \ \dfrac{1}{6} [/tex]