Nierówności logarytmiczne

[Tupek]

Niby proste ale jednak nie wiem jak rozwiazać:

1. \(\displaystyle{ log_{\frac{1}{4}}(x+1)-2log_{\frac{1}{4}}x>-\frac{1}{2}}\)

2. \(\displaystyle{ log_{\frac{1}{3}}[log_{4}(x-5)]>0}\)

3. \(\displaystyle{ log_{2}[log_{\frac{1}{2}}(x-4)]}\)

[siwy]

1
\(\displaystyle{ log_{\frac{1}{4}} (x+1)-log_{\frac{1}{4}} x \rangle - \frac{1}{2} \\ log_{\frac{1}{4}} \frac{x+1}{x} \rangle log_{\frac{1}{4}} 2 \\ \frac{x+1}{x} \langle 2 \\ x+1 \langle 2x \\ x \rangle 1}\)

[ Dodano: Nie Lis 13, 2005 3:04 pm ]
2
\(\displaystyle{ log_{\frac{1}{3}}[log_{4}(x-5)] \rangle 0 = log_{\frac{1}{3}}1 \\ log_{4} (x-5) \langle 1 = log_{4} 4 \\ x-5 \langle 4 \\ x \langle 9}\)
ale
\(\displaystyle{ x-5 \rangle 0 \\ x \rangle 5 \\ i \\ log_{4}(x-5) \rangle 0 \\ wiec \\ x-5 \rangle 1 \\ x \rangle 6}\)
łącznie
\(\displaystyle{ x\in (6,9)}\)

[ Dodano: Nie Lis 13, 2005 3:15 pm ]
3
\(\displaystyle{ log_{2}[log_{\frac{1}{2}}(x-4)] \langle 0 = log_{2}1 \\ log_{\frac{1}{2}}(x-4) \langle 1 = log_{\frac{1}{2}}\frac{1}{2} \\ x-4 \rangle \frac{1}{2} \\ x \rangle \frac{9}{2}}\)
oraz
\(\displaystyle{ x-4 \rangle 0 \\ x \rangle 4}\)
i
\(\displaystyle{ log_{\frac{1}{2}}(x-4) \rangle 0 \\ x-4 \langle 1 \\ x \langle 5}\)
łącznie
\(\displaystyle{ x (\frac{9}{2},5)}\)