Question

Fie a, b>1. Demonstrati ca $log_{a}(b+1)\ \textgreater \ log_{a+1}(b+1)\ \textgreater \ log_{a+1}b$. Mersi

Answer 1

$Pentru\ x\ \textgreater \ 1\ functia\ \log_xk\ este\ strict\ descrescatoare,\ adica \\ \log_xk\ \textgreater \ log_{x+\alpha}k\ \forall\ \alpha\ \textgreater \ 0.\ Deci\ \log_a(b+1)\ \textgreater \ \log_{a+1}(b+1). \\ Functia\ \log_kx\ este\ strict\ crescatoare\ pentru\ k\ \textgreater \ 1;x\ \textgreater \ 1. \\ Deci\ \log_{a+1}(b+1)\ \textgreater \ \log_{a+1}b$

Answer 2

a, b > 1

[tex]\log_a(b+1) \ \textgreater \ \dfrac{\log_a(b+1)}{\log_a(a+1)}\ \textgreater \ \dfrac{\log_ab}{\log_a(a+1)} \\\;\\ \log_a (a+1) \ \textgreater \ \log_a a =1 [/tex]