#bobhans# 3^x −2^(x+1) ≤ (√(2.9^x −10.6^x +2^(2x+3) )) find the solution set


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Question Number 106695 by bobhans last updated on 06/Aug/20

$You can't use 'macro parameter character #' in math mode$ $3^{x} - 2^{x + 1} ⩽ \sqrt{2 . 9^{x} - 10 . 6^{x} + 2^{2 x + 3}}$ $find the solution set$
	Answered by bemath last updated on 06/Aug/20
	$^(@bemath@) ((3^x −2.2^x )/2^x ) ≤ ((√(2.9^x −10.6^x +8.(2^x )^2 ))/2^x ) ((3/2))^x −2 ≤ (√(2((3/2))^(2x) −10((3/2))^x +8)) set ((3/2))^x = t t−2 ≤ (√(2(t−1)(t−4))) defined on (t−1)(t−4)≥0 ⇒ t ≤1 ∪ t ≥ 4...(1) squaring both sides t^2 −4t+4 ≤ 2t^2 −10t+8 −t^2 +6t−4 ≤ 0 ; t^2 −6t+4 ≥ 0 (t−3)^2 −5≥0 t≤3−(√5) ∪t ≥ 3+(√5) ...(2) from (1)∩(2) we get → { ((t ≤1)),((t≥3+(√5))) :} { ((((3/2))^x ≤1 ⇒ x ≤ 0)),((((3/2))^x ≥ 3+(√5) ⇒x ≥log _(((3/2))) (3+(√5)) )) :} the solution set is x∈(−∞,0 ] ∪ [ log _(((3/2))) (3+(√5)) ,∞ )$
	$\overset{@ bemath @}{}$ $\frac{3^{x} - 2 . 2^{x}}{2^{x}} ⩽ \frac{\sqrt{2 . 9^{x} - 10 . 6^{x} + 8 . {(2^{x})}^{2}}}{2^{x}}$ ${(\frac{3}{2})}^{x} - 2 ⩽ \sqrt{2 {(\frac{3}{2})}^{2 x} - 10 {(\frac{3}{2})}^{x} + 8}$ $set {(\frac{3}{2})}^{x} = t$ $t - 2 ⩽ \sqrt{2 (t - 1) (t - 4)}$ $defined on (t - 1) (t - 4) ⩾ 0$ $\Rightarrow t ⩽ 1 \cup t ⩾ 4 . . . (1)$ $squaring both sides$ $t^{2} - 4 t + 4 ⩽ {2 t}^{2} - 10 t + 8$ $- t^{2} + 6 t - 4 ⩽ 0; t^{2} - 6 t + 4 ⩾ 0$ ${(t - 3)}^{2} - 5 ⩾ 0$ $t ⩽ 3 - \sqrt{5} \cup t ⩾ 3 + \sqrt{5} . . . (2)$ $from (1) \cap (2) we get \to {\begin{cases} t ⩽ 1 \\ t ⩾ 3 + \sqrt{5} \end{cases}$ ${\begin{cases} {(\frac{3}{2})}^{x} ⩽ 1 \Rightarrow x ⩽ 0 \\ {(\frac{3}{2})}^{x} ⩾ 3 + \sqrt{5} \Rightarrow x ⩾ \log_{(\frac{3}{2})} (3 + \sqrt{5}) \end{cases}$ $the solution set is x \in (- \infty, 0] \cup$ $[\log_{(\frac{3}{2})} (3 + \sqrt{5}), \infty)$
	Commented by bobhans last updated on 06/Aug/20

	$afdollll$
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