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Question Number 165178 by cortano1 last updated on 27/Jan/22

$$x\in R\Rightarrow \mid \log {}_2\left(\frac{x}{2}\right){\mid }^3+\mid \log {}_2\left(2x\right){\mid }^3=28$$

Answered by aleks041103 last updated on 27/Jan/22

$${log}_2\left(x/2\right)={log}_{2}x-1$$$${log}_2\left(2x\right)={log}_{2}x+1$$$$\Rightarrow lett={log}_{2}x$$$$\mid t+1{\mid }^3+\mid t-1{\mid }^3=28=f\left(t\right)$$$$f(-t)=\mid 1-t{\mid }^3+\mid -t-1{\mid }^3=$$$$=\mid t+1{\mid }^3+\mid t-1{\mid }^3=f\left(t\right)$$$$f\left(t\right)iseven$$$$$$$$1)t⩽-1:$$$$\mid t+1\mid =-t-1$$$$\mid t-1\mid =1-t$$$$\Rightarrow {(1-t)}^3-{(1+t)}^3=28$$$$-2t(1-2t+t^2+1+2t+t^2+1-t^2)=28$$$$-2t(3+t^2)=28$$$$-2t^3-6t=28$$$$p\left(t\right)=t^3+3t+14=0$$$$p^{\prime }\left(t\right)=3t^2+3>0$$$$\Rightarrow p\left(t\right)isstrictlyincreasing$$$$\Rightarrow hasatmost1root.$$$$byobs.t=-2isasol.andalsot=-2⩽-1.✓$$$$$$$$2)0⩾t⩾-1:$$$$\mid t+1\mid =t+1$$$$\mid t-1\mid =1-t$$$${(1+t)}^3+{(1-t)}^3=28$$$$2(1+2t+t^2+1-2t+t^2-1+t^2)=28$$$$3t^2-13=0$$$$\Rightarrow t=\pm \sqrt{\frac{13}{3}}\notin [-1,0]$$$$\Rightarrow nosol.in[-1,0].$$$$$$$$f\left(t\right)iseven\Rightarrow allsols.aret=\pm 2$$$$\Rightarrow {log}_{2}x=\pm 2$$$$\Rightarrow x=2^{\pm 2}$$$$\Rightarrow x=\frac{1}{4}andx=4$$

Answered by MJS_new last updated on 27/Jan/22

$$x=2^t$$$$\mid t-1{\mid }^3+\mid t+1{\mid }^3=28$$$$\left(1\right)t⩽-1$$$$t^3+3t+14=0\Rightarrow t=-2$$$$\left(2\right)-1<t⩽1$$$$t^2-\frac{13}{3}=0\Rightarrow \mathrm{no}\mathrm{solution}$$$$\left(3\right)1<t$$$$t^3+3t-14=0\Rightarrow t=2$$$$\Rightarrow$$$$x=\frac{1}{4}\vee x=4$$

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