19 uždavinys
Sprendimas.
2^(5-x^2) ≤
16
16
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2^(5-x^2) ≤ 16$$2^{(5-x^2)}$$ ≤ $$16$$
log(2, 2^(5-x^2)) ≤ log(2,16)$$log_{2}(2^{(5-x^2)})$$ ≤ $$log_{2}(16)$$
$${\normalsize log_{2}(16)}$$ = $${\normalsize 4}$$
log(2, 2^(5-x^2)) ≤ 4$$log_{2}(2^{(5-x^2)})$$ ≤ $$4$$
(5- x^2) ≤ 4$$(5-x^{2})$$ ≤ $$4$$
$${\normalsize (5-x^{2})}$$ = $${\normalsize 5-x^{2}}$$
5- x^2 ≤ 4$$5-x^{2}$$ ≤ $$4$$
5 ≤ 4+ x^2$$5$$ ≤ $$4+x^{2}$$
5-4 ≤ x^2$$5-4$$ ≤ $$x^{2}$$
$${\normalsize 5-4}$$ = $${\normalsize 1}$$
1 ≤ x^2$$1$$ ≤ $$x^{2}$$
x^2 ≥ 1$$x^{2}$$ ≥ $$1$$
$$2^{(5-x^2)}$$ ≤ $$16$$
$$log_{2}(2^{(5-x^2)})$$ ≤ $$log_{2}(16)$$
$$log_{2}(2^{(5-x^2)})$$ ≤ $$4$$
$$5-x^{2}$$ ≤ $$4$$
$$5-4$$ ≤ $$x^{2}$$
$$1$$ ≤ $$x^{2}$$
$$x^{2}$$ ≥ $$1$$
x priklauso [-∞; -1] U [ 1; +∞]
Atsakymas: x priklauso [-∞; -1] U [ 1; +∞]