Question

Решите неравенства log1/2(x+3)>-2

Answer 1

$log_{frac{1}{2}}(x 3)>-2\ log_{2^{-1}}(x 3)>-2\ -log_{2}(x 3)>-2\ log_{2}(x 3)<2\ log_{2}(x 3)<log_{2}4$</var></p> <p><img src=[/tex]right.\ left {{{x+3<4} atop {x+3>0}} right.\" title="log_{frac{1}{2}}(x+3)>-2\ log_{2^{-1}}(x+3)>-2\ -log_{2}(x+3)>-2\ log_{2}(x+3)<2\ log_{2}(x+3)<log_{2}4" title="right.\ left {{{x+3<4} atop {x+3>0}} right.\" title="log_{frac{1}{2}}(x+3)>-2\ log_{2^{-1}}(x+3)>-2\ -log_{2}(x+3)>-2\ log_{2}(x+3)<2\ log_{2}(x+3)<log_{2}4" alt="right.\ left {{{x+3<4} atop {x+3>0}} right.\" title="log_{frac{1}{2}}(x+3)>-2\ log_{2^{-1}}(x+3)>-2\ -log_{2}(x+3)>-2\ log_{2}(x+3)<2\ log_{2}(x+3)<log_{2}4" />

$log_{frac{1}{2}}(x+3)>-2\ log_{2^{-1}}(x+3)>-2\ -log_{2}(x+3)>-2\ log_{2}(x+3)<2\ log_{2}(x+3)<log_{2}4$

$<var>right.\ left {{{x 3<4} atop {x 3>0}} right.$

[tex]right.\ left {{{x<1} atop {x>-3}} right.\" title="right.\ left {{{x+3<4} atop {x+3>0}} right.\" />

[tex]right.\ left {{{x<1} atop {x>-3}} right.\" alt="right.\ left {{{x+3<4} atop {x+3>0}} right.\" />

[tex]right.\ left {{{x<1} atop {x>-3}} right.\" />

x принадлежит интервалу (-3;1)