Question Number 165178 by cortano1 last updated on 27/Jan/22
$$\:\:{x}\in{R}\:\Rightarrow\:\mid\:\mathrm{log}\:_{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)\mid^{\mathrm{3}} +\mid\mathrm{log}\:_{\mathrm{2}} \left(\mathrm{2}{x}\right)\mid^{\mathrm{3}} =\mathrm{28} \\ $$
Answered by aleks041103 last updated on 27/Jan/22
![log_2 (x/2)=log_2 x−1 log_2 (2x)=log_2 x+1 ⇒let t=log_2 x ∣t+1∣^3 +∣t−1∣^3 =28=f(t) f(−t)=∣1−t∣^3 +∣−t−1∣^3 = =∣t+1∣^3 +∣t−1∣^3 =f(t) f(t) is even 1)t≤−1: ∣t+1∣=−t−1 ∣t−1∣=1−t ⇒(1−t)^3 −(1+t)^3 =28 −2t(1−2t+t^2 +1+2t+t^2 +1−t^2 )=28 −2t(3+t^2 )=28 −2t^3 −6t=28 p(t)=t^3 +3t+14=0 p′(t)=3t^2 +3>0 ⇒p(t) is strictly increasing ⇒ has at most 1 root. by obs. t=−2 is a sol. and also t=−2≤−1.✓ 2)0≥t≥−1: ∣t+1∣=t+1 ∣t−1∣=1−t (1+t)^3 +(1−t)^3 =28 2(1+2t+t^2 +1−2t+t^2 −1+t^2 )=28 3t^2 −13=0 ⇒t=±(√((13)/3))∉[−1,0] ⇒ no sol. in [−1,0]. f(t) is even ⇒ all sols. are t=±2 ⇒log_2 x=±2 ⇒x=2^(±2) ⇒x=(1/4) and x=4](https://www.tinkutara.com/question/Q165182.png)
$${log}_{\mathrm{2}} \left({x}/\mathrm{2}\right)={log}_{\mathrm{2}} {x}−\mathrm{1} \\ $$$${log}_{\mathrm{2}} \left(\mathrm{2}{x}\right)={log}_{\mathrm{2}} {x}+\mathrm{1} \\ $$$$\Rightarrow{let}\:{t}={log}_{\mathrm{2}} {x} \\ $$$$\mid{t}+\mathrm{1}\mid^{\mathrm{3}} +\mid{t}−\mathrm{1}\mid^{\mathrm{3}} =\mathrm{28}={f}\left({t}\right) \\ $$$${f}\left(−{t}\right)=\mid\mathrm{1}−{t}\mid^{\mathrm{3}} +\mid−{t}−\mathrm{1}\mid^{\mathrm{3}} = \\ $$$$=\mid{t}+\mathrm{1}\mid^{\mathrm{3}} +\mid{t}−\mathrm{1}\mid^{\mathrm{3}} ={f}\left({t}\right) \\ $$$${f}\left({t}\right)\:{is}\:{even} \\ $$$$ \\ $$$$\left.\mathrm{1}\right){t}\leqslant−\mathrm{1}: \\ $$$$\mid{t}+\mathrm{1}\mid=−{t}−\mathrm{1} \\ $$$$\mid{t}−\mathrm{1}\mid=\mathrm{1}−{t} \\ $$$$\Rightarrow\left(\mathrm{1}−{t}\right)^{\mathrm{3}} −\left(\mathrm{1}+{t}\right)^{\mathrm{3}} =\mathrm{28} \\ $$$$−\mathrm{2}{t}\left(\mathrm{1}−\mathrm{2}{t}+{t}^{\mathrm{2}} +\mathrm{1}+\mathrm{2}{t}+{t}^{\mathrm{2}} +\mathrm{1}−{t}^{\mathrm{2}} \right)=\mathrm{28} \\ $$$$−\mathrm{2}{t}\left(\mathrm{3}+{t}^{\mathrm{2}} \right)=\mathrm{28} \\ $$$$−\mathrm{2}{t}^{\mathrm{3}} −\mathrm{6}{t}=\mathrm{28} \\ $$$${p}\left({t}\right)={t}^{\mathrm{3}} +\mathrm{3}{t}+\mathrm{14}=\mathrm{0} \\ $$$${p}'\left({t}\right)=\mathrm{3}{t}^{\mathrm{2}} +\mathrm{3}>\mathrm{0} \\ $$$$\Rightarrow{p}\left({t}\right)\:{is}\:{strictly}\:{increasing} \\ $$$$\Rightarrow\:{has}\:{at}\:{most}\:\mathrm{1}\:{root}. \\ $$$${by}\:{obs}.\:{t}=−\mathrm{2}\:{is}\:{a}\:{sol}.\:{and}\:{also}\:{t}=−\mathrm{2}\leqslant−\mathrm{1}.\checkmark \\ $$$$ \\ $$$$\left.\mathrm{2}\right)\mathrm{0}\geqslant{t}\geqslant−\mathrm{1}: \\ $$$$\mid{t}+\mathrm{1}\mid={t}+\mathrm{1} \\ $$$$\mid{t}−\mathrm{1}\mid=\mathrm{1}−{t} \\ $$$$\left(\mathrm{1}+{t}\right)^{\mathrm{3}} +\left(\mathrm{1}−{t}\right)^{\mathrm{3}} =\mathrm{28} \\ $$$$\mathrm{2}\left(\mathrm{1}+\mathrm{2}{t}+{t}^{\mathrm{2}} +\mathrm{1}−\mathrm{2}{t}+{t}^{\mathrm{2}} −\mathrm{1}+{t}^{\mathrm{2}} \right)=\mathrm{28} \\ $$$$\mathrm{3}{t}^{\mathrm{2}} −\mathrm{13}=\mathrm{0} \\ $$$$\Rightarrow{t}=\pm\sqrt{\frac{\mathrm{13}}{\mathrm{3}}}\notin\left[−\mathrm{1},\mathrm{0}\right] \\ $$$$\Rightarrow\:{no}\:{sol}.\:{in}\:\left[−\mathrm{1},\mathrm{0}\right]. \\ $$$$ \\ $$$${f}\left({t}\right)\:{is}\:{even}\:\Rightarrow\:{all}\:{sols}.\:{are}\:{t}=\pm\mathrm{2} \\ $$$$\Rightarrow{log}_{\mathrm{2}} {x}=\pm\mathrm{2} \\ $$$$\Rightarrow{x}=\mathrm{2}^{\pm\mathrm{2}} \\ $$$$\Rightarrow{x}=\frac{\mathrm{1}}{\mathrm{4}}\:{and}\:{x}=\mathrm{4} \\ $$
Answered by MJS_new last updated on 27/Jan/22
$${x}=\mathrm{2}^{{t}} \\ $$$$\mid{t}−\mathrm{1}\mid^{\mathrm{3}} +\mid{t}+\mathrm{1}\mid^{\mathrm{3}} =\mathrm{28} \\ $$$$\left(\mathrm{1}\right)\:{t}\leqslant−\mathrm{1} \\ $$$$\:\:\:\:\:{t}^{\mathrm{3}} +\mathrm{3}{t}+\mathrm{14}=\mathrm{0}\:\Rightarrow\:{t}=−\mathrm{2} \\ $$$$\left(\mathrm{2}\right)\:−\mathrm{1}<{t}\leqslant\mathrm{1} \\ $$$$\:\:\:\:\:{t}^{\mathrm{2}} −\frac{\mathrm{13}}{\mathrm{3}}=\mathrm{0}\:\Rightarrow\:\mathrm{no}\:\mathrm{solution} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{1}<{t} \\ $$$$\:\:\:\:\:{t}^{\mathrm{3}} +\mathrm{3}{t}−\mathrm{14}=\mathrm{0}\:\Rightarrow\:{t}=\mathrm{2} \\ $$$$\Rightarrow \\ $$$${x}=\frac{\mathrm{1}}{\mathrm{4}}\vee{x}=\mathrm{4} \\ $$