Question Number 106695 by bobhans last updated on 06/Aug/20
$$\:\:\:\:#\mathrm{bobhans}# \\ $$$$\mathrm{3}^{\mathrm{x}} −\mathrm{2}^{\mathrm{x}+\mathrm{1}} \:\leqslant\:\sqrt{\mathrm{2}.\mathrm{9}^{\mathrm{x}} −\mathrm{10}.\mathrm{6}^{\mathrm{x}} +\mathrm{2}^{\mathrm{2x}+\mathrm{3}} } \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{solution}\:\mathrm{set} \\ $$
Answered by bemath last updated on 06/Aug/20
![^(@bemath@) ((3^x −2.2^x )/2^x ) ≤ ((√(2.9^x −10.6^x +8.(2^x )^2 ))/2^x ) ((3/2))^x −2 ≤ (√(2((3/2))^(2x) −10((3/2))^x +8)) set ((3/2))^x = t t−2 ≤ (√(2(t−1)(t−4))) defined on (t−1)(t−4)≥0 ⇒ t ≤1 ∪ t ≥ 4...(1) squaring both sides t^2 −4t+4 ≤ 2t^2 −10t+8 −t^2 +6t−4 ≤ 0 ; t^2 −6t+4 ≥ 0 (t−3)^2 −5≥0 t≤3−(√5) ∪t ≥ 3+(√5) ...(2) from (1)∩(2) we get → { ((t ≤1)),((t≥3+(√5))) :} { ((((3/2))^x ≤1 ⇒ x ≤ 0)),((((3/2))^x ≥ 3+(√5) ⇒x ≥log _(((3/2))) (3+(√5)) )) :} the solution set is x∈(−∞,0 ] ∪ [ log _(((3/2))) (3+(√5)) ,∞ )](https://www.tinkutara.com/question/Q106697.png)
$$\:\:\:\:\:\:\:\overset{@\mathrm{bemath}@} {\:} \\ $$$$\frac{\mathrm{3}^{\mathrm{x}} −\mathrm{2}.\mathrm{2}^{\mathrm{x}} }{\mathrm{2}^{\mathrm{x}} }\:\leqslant\:\frac{\sqrt{\mathrm{2}.\mathrm{9}^{\mathrm{x}} −\mathrm{10}.\mathrm{6}^{\mathrm{x}} +\mathrm{8}.\left(\mathrm{2}^{\mathrm{x}} \right)^{\mathrm{2}} }}{\mathrm{2}^{\mathrm{x}} } \\ $$$$\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{x}} −\mathrm{2}\:\leqslant\:\sqrt{\mathrm{2}\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2x}} −\mathrm{10}\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{x}} +\mathrm{8}} \\ $$$$\mathrm{set}\:\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{x}} =\:\mathrm{t}\: \\ $$$$\mathrm{t}−\mathrm{2}\:\leqslant\:\sqrt{\mathrm{2}\left(\mathrm{t}−\mathrm{1}\right)\left(\mathrm{t}−\mathrm{4}\right)} \\ $$$$\mathrm{defined}\:\mathrm{on}\:\left(\mathrm{t}−\mathrm{1}\right)\left(\mathrm{t}−\mathrm{4}\right)\geqslant\mathrm{0} \\ $$$$\Rightarrow\:\mathrm{t}\:\leqslant\mathrm{1}\:\cup\:\mathrm{t}\:\geqslant\:\mathrm{4}…\left(\mathrm{1}\right) \\ $$$$\mathrm{squaring}\:\mathrm{both}\:\mathrm{sides} \\ $$$$\mathrm{t}^{\mathrm{2}} −\mathrm{4t}+\mathrm{4}\:\leqslant\:\mathrm{2t}^{\mathrm{2}} −\mathrm{10t}+\mathrm{8} \\ $$$$−\mathrm{t}^{\mathrm{2}} +\mathrm{6t}−\mathrm{4}\:\leqslant\:\mathrm{0}\:;\:\mathrm{t}^{\mathrm{2}} −\mathrm{6t}+\mathrm{4}\:\geqslant\:\mathrm{0} \\ $$$$\left(\mathrm{t}−\mathrm{3}\right)^{\mathrm{2}} −\mathrm{5}\geqslant\mathrm{0} \\ $$$$\mathrm{t}\leqslant\mathrm{3}−\sqrt{\mathrm{5}}\:\cup\mathrm{t}\:\geqslant\:\mathrm{3}+\sqrt{\mathrm{5}}\:…\left(\mathrm{2}\right) \\ $$$$\mathrm{from}\:\left(\mathrm{1}\right)\cap\left(\mathrm{2}\right)\:\mathrm{we}\:\mathrm{get}\:\rightarrow\begin{cases}{\mathrm{t}\:\leqslant\mathrm{1}}\\{\mathrm{t}\geqslant\mathrm{3}+\sqrt{\mathrm{5}}}\end{cases} \\ $$$$\begin{cases}{\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{x}} \leqslant\mathrm{1}\:\Rightarrow\:\mathrm{x}\:\leqslant\:\mathrm{0}}\\{\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{x}} \geqslant\:\mathrm{3}+\sqrt{\mathrm{5}}\:\Rightarrow\mathrm{x}\:\geqslant\mathrm{log}\:_{\left(\frac{\mathrm{3}}{\mathrm{2}}\right)} \left(\mathrm{3}+\sqrt{\mathrm{5}}\right)\:}\end{cases} \\ $$$$\mathrm{the}\:\mathrm{solution}\:\mathrm{set}\:\mathrm{is}\:\mathrm{x}\in\left(−\infty,\mathrm{0}\:\right]\:\cup \\ $$$$\left[\:\mathrm{log}\:_{\left(\frac{\mathrm{3}}{\mathrm{2}}\right)} \left(\mathrm{3}+\sqrt{\mathrm{5}}\right)\:,\infty\:\right)\: \\ $$$$ \\ $$
Commented by bobhans last updated on 06/Aug/20
$$\mathrm{afdollll} \\ $$