Nierówności logarytmiczne

[-Klaudia-]

Rozwiąż nierówności:
a) \(\3^{-x^3} \cdot 9 < (\frac {\sqrt3} {3}) ^{6x-4}\)
b) \(\log_x [log_2(4^x - 6)] =1\)
c) \(\log(\frac{|x|}{x} + x) = 1\)

[domino21]

a.
\(3^{-x^3} \cdot 9 <(\frac{\sqrt{3}}{3})^{6x-4}
3^{-x^3} \cdot 3^2 < 3^{-\frac{1}{2}(6x-4)}
3^{-x^3+2}< 3^{-3x+2}
-x^3+2<-3x+2
x^3-3x>0
x(x^2-3)>0
x(x-\sqrt{3})(x+\sqrt{3})>0
x\in (-\sqrt{3},0) \cup (\sqrt{3};+\infty)\)

[domino21]

b.
\(\log_x [ \log_2(4^x -6)]=1, \ \ D: \ x>0, \ x\neq 1
\log_x [ \log_2(4^x-6)]=\log_x x
\log_2(4^x -6) =x
\log_2 (4^x-6)=\log_2 2^x
4^x-6=2^x
2^{2x}-2^x-6=0\)

\(2^x =t \ \wedge \ t>0
t^2-t-6=0
(t-3)(t+2)=0
t=3 \ \vee \ t=-2 \ \wedge \ t>0
t=3\)

\(2^x =3
\log_2 2^x =\log_2 3
x=\log_2 3\)

[domino21]

c.
\(\log(\frac{|x|}{x}+x)=1
\log(\frac{|x|}{x}+x)=\log 10
\frac{|x|}{x} +x =10
|x|+x^2=10x\)

\(1^{\circ} \ x>0
x+x^2=10x \ \Rightarrow \ x^2-9x=0 \ \Rightarrow \ x(x-9)=0 \ \wedge \ x>0 \ \Rightarrow \ x=9\)

\(2^{\circ} \ x<0
-x+x^2=10x \ \Rightarrow \ x^2-11x=0 \ \Rightarrow \ x(x-11)=0 \ \wedge \ x<0 \ \Rightarrow \ x\in \empty\)

\(D:
\frac{|x|}{x}+x>0
\frac{|x|+x}{x}>0
x(|x|+x)>0\)

\(1^{\circ} \ x>0
x(x+x)>0 \ \Rightarrow \ 3x>0 \ \wedge \ x>0 \ \Rightarrow \ x>0\)

\(2^{\circ} \ x<0
x(-x+x)<0 \ \Rightarrow \ 0<0 \ \Rightarrow \ x\in \empty\)

\(D=(0;+\infty) \ \wedge \ x=9 \in D\)