Bonjour IpekTufan
[tex]1)\ \log_{2^{-1}}(2x+1)\ \textless \ 3\\\\Condition:2x+1\ \textgreater \ 0\Longrightarrow x\ \textgreater \ -\dfrac{1}{2}\\\\\dfrac{\ln(2x+1)}{\ln2^{-1}}\ \textless \ 3\\\\-\dfrac{\ln(2x+1)}{\ln2}\ \textless \ 3\\\\-\ln(2x+1)\ \textless \ 3\ln2\\\\-\ln(2x+1)\ \textless \ \ln2^3\\\\-\ln(2x+1)\ \textless \ \ln8\\\\\ln8+\ln(2x+1)\ \textgreater \ 0\\\\\ln8(2x+1)\ \textgreater \ 0\\\\\ln8(2x+1)\ \textgreater \ \ln1\\\\8(2x+1)\ \textgreater \ 1\\\\16x+8\ \textgreater \ 1\\\\16x\ \textgreater \ -7\\\\x\ \textgreater \ -\dfrac{7}{16}[/tex]
La condition est vérifiée.
Par conséquent, l'ensemble des solutions de l'inéquation est [tex]\boxed{S=]-\dfrac{7}{16};+\infty[}[/tex]
[tex]2)\ \log_{\frac{1}{5}}(x-2)\ \textgreater \ 1\\\\Condition:x-2\ \textgreater \ 0\Longrightarrow x\ \textgreater \ 2\\\\\dfrac{\ln(x-2)}{\ln\frac{1}{5}}\ \textgreater \ 1\\\\\\-\dfrac{\ln(x-2)}{\ln5}\ \textgreater \ 1\\\\-\ln(x-2)\ \textgreater \ \ln5\\\\\ln5+\ln(x-2)\ \textless \ 0\\\\\ln5(x-2)\ \textless \ 0\\\\\ln5(x-2)\ \textless \ \ln1\\\\5(x-2)\ \textless \ 1\\\\5x-10\ \textless \ 1\\\\5x\ \textless \ 11\\\\x\ \textless \ \dfrac{11}{5}\\\\Or\ x\ \textless \ 2\\\\Donc\ 2\ \textless \ x\ \textless \ \dfrac{11}{5}[/tex]
Par conséquent, l'ensemble des solutions de l'inéquation est [tex]\boxed{S=]2;\dfrac{11}{5}[}[/tex]
[tex]3)\ \log_{\frac{1}{3}}(x-\dfrac{2}{9})\ \textless \ 2\\\\Condition:x-\dfrac{2}{9}\ \textgreater \ 0\Longrightarrow x\ \textgreater \ \dfrac{2}{9}\\\\\dfrac{\ln(x-\dfrac{2}{9})}{\ln\frac{1}{3}}\ \textless \ 2\\\\-\dfrac{\ln(x-\dfrac{2}{9})}{\ln3}\ \textless \ 2\\\\-\ln(x-\dfrac{2}{9})\ \textless \ 2\ln3\\\\-\ln(x-\dfrac{2}{9})\ \textless \ \ln3^2\\\\-\ln(x-\dfrac{2}{9})\ \textless \ \ln9\\\\\ln9+\ln(x-\dfrac{2}{9})\ \textgreater \ 0\\\\\ln9(x-\dfrac{2}{9})\ \textgreater \ 0\\\\\ln(9x-2)\ \textgreater \ 0\\\\\ln(9x-2)\ \textgreater \ \ln1\\\\9x-2\ \textgreater \ 1\\\\9x\ \textgreater \ 3\\\\x\ \textgreater \ \dfrac{3}{9}\\\\x\ \textgreater \ \dfrac{1}{3}[/tex]
La condition est vérifiée.
Par conséquent, l'ensemble des solutions de l'inéquation est [tex]\boxed{S=]\dfrac{1}{3};+\infty[}[/tex]
[tex]4)\ \log_{\frac{1}{16}}(x+1)\ \textgreater \ \dfrac{1}{4}\\\\Condition:x+1\ \textgreater \ 0\Longrightarrow x\ \textgreater \ -1\\\\\dfrac{\ln(x+1)}{\ln\frac{1}{16}} \ \textgreater \ \dfrac{1}{4}\\\\-\dfrac{\ln(x+1)}{\ln16} \ \textgreater \ \dfrac{1}{4}\\\\-\dfrac{\ln(x+1)}{\ln2^4} \ \textgreater \ \dfrac{1}{4}\\\\-\dfrac{\ln(x+1)}{4\ln2} \ \textgreater \ \dfrac{1}{4}\\\\-\dfrac{\ln(x+1)}{\ln2} \ \textgreater \ 1\\\\-\ln(x+1)\ \textgreater \ \ln2\\\\\ln2+\ln(x+1)\ \textless \ 0\\\\\ln2(x+1)\ \textless \ \ln1\\\\2(x+1)\ \textless \ 1\\\\2x+2\ \textless \ 1\\\\2x\ \textless \ -1\\\\x\ \textless \ -\dfrac{1}{2}[/tex]
Or x > -1
Donc [tex]-1\ \textless \ x\ \textless \ -\dfrac{1}{2}[/tex]
Par conséquent, l'ensemble des solutions de l'inéquation est [tex]\boxed{S=]-1;-\dfrac{1}{2}[}[/tex]
