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Question

The solution set of $$ \log _{ 1/2 }{ \left( { 2 }^{ x+2 }-{ 4 }^{ x } \right)  } \ge -2 $$  is


Solution

$$\log_{y_2}(2^{x+2}-4^x) \ge -2$$
$$\therefore \ -\log_2 (2^{x+2}-4^x) \ge -2$$
$$\therefore \ -\log_2 (2^{x+2}-4^x) \le -2$$
$$\therefore \ 2^{x+2}-4^x \le 4$$
$$\therefore \ 4.x2^2-2^{2x} \le 4$$
Let $$2^x =t$$
$$\therefore \ 4t-t^2$$
$$\therefore \ 0 \le t^2-4t+4$$
$$\therefore \ 0 \le (t-2)^2$$
$$H$$ values


Mathematics

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