Question

$\\ Aratati ca: \\ \\ \Big(\log_\big{6} 12 - \log_{\big{12}}24\Big)^{-1} \in \big(8,10\big).$

Answer 1

Sper sa te ajute...nu stiu alt mod de a calcula log in baza 2 din 3 :))

Answer 2

$\displaystyle Voi~nota~cu~E~valoarea~expresiei~si~cu~F=E^{-1}.~(Deci~F~este \\ \\~expresia~din~paranteza~ paranteza) \\ \\ Notam~t= \log_23. \\ \\ Avem~F= \log_62+1- \log_{12}2-1= \log_62- \log_{12}2= \\ \\ = \frac{1}{\log_26}- \frac{1}{\log_2{12}}= \frac{1}{t+1}- \frac{1}{t+2}= \frac{1}{(t+1)(t+2)}. \\ \\ Deci~ \boxed{E=(t+1)(t+2)}~.$

$\displaystyle Am~vazut~in~multe~lucrari~urmatoarea~problema \\ \\ "Demonstrati~ca~ \log_23 \in (1,5;1,6)." \\ \\ (de~exemplu~"Exponentiale~si~logaritmi"~(Gheorghe~Andrei) \\ \\ si~intr-un~supliment~al~Gazetei~Matematice).~Am~vazut~de \\ \\ asemenea~probleme~care~se~rezolvau~cu~acest~rezultat. \\ \\ Sa~demonstram~aceasta~afirmatie!$

$\displaystyle \log_2{3}\ \textgreater \ 1,5 \Leftrightarrow 2 \log_23\ \textgreater \ 3 \Leftrightarrow \log_29\ \textgreater \ \log_28,~adevarat! \\ \\ \log_23\ \textless \ 1,6 \Leftrightarrow 5 \log_23\ \textless \ 8 \Leftrightarrow \log_2243\ \textless \ \log_2256,~adevarat!$

$\displaystyle~Revenind~la~problema~noastra...~Am~demonstrat~ca~t \in (1,5~;~1,6). \\ \\ Ne-a~ramas~de~demonstrat~ca~f(t)=(t+1)(t+2) \in (8,10). \\ \\ f:(0, + \infty) \to \mathbb{R}~-~strict~crescatoare. \\ \\t\ \textgreater \ 1,5 \Rightarrow f(t)\ \textgreater \ f(1,5)=2,5 \cdot 3,5=8,75. \\ \\ t\ \textless \ 1,6 \Rightarrow f(t)\ \textless \ f(1,6)=2,6 \cdot 3,6=9,36. \\ \\ Deci~E \in (8,75~;~9,36) \subset (8,10).$