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Question

Solve the following inequality:
$$\displaystyle\, \log_{1/3}\, (2^{x + 2} - 4^x)\, \geqslant \, -2$$


Solution

$$let\quad { 2 }^{ x }=z$$
$$ -\log _{ 3 }{ (4z-{ z }^{ 2 }) } \ge -2$$
$$ 4z-{ z }^{ 2 }\le 9$$
$$ { z }^{ 2 }-4z+9\ge 0$$
$$ therefore\quad z\quad \varepsilon R$$
$$ now,{ 2 }^{ x+2 }-{ 4 }^{ x }>0$$
$$ as\quad { 2 }^{ x }=z$$
$$ 4z-{ z }^{ 2 }>0$$
$$ { z }^{ 2 }-4z<0$$
$$ z(z-4)<0$$
$$ z\quad \varepsilon (0,4)$$
$$ { 2 }^{ x }\quad \varepsilon (0,4)$$
$$ x\quad \varepsilon (-\infty ,2)$$

Maths

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