Menu Close

x-R-log-2-x-2-3-log-2-2x-3-28-

Tinku Tara 0 Comments
Pin




Question Number 165178 by cortano1 last updated on 27/Jan/22
x∈R ⇒ ∣ log _2 ((x/2))∣^3 +∣log _2 (2x)∣^3 =28
x∈R⇒∣log2(x2)∣3+∣log2(2x)∣3=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
log2(x/2)=log2x−1log2(2x)=log2x+1⇒lett=log2x∣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)iseven1)t⩽−1:∣t+1∣=−t−1∣t−1∣=1−t⇒(1−t)3−(1+t)3=28−2t(1−2t+t2+1+2t+t2+1−t2)=28−2t(3+t2)=28−2t3−6t=28p(t)=t3+3t+14=0p′(t)=3t2+3>0⇒p(t)isstrictlyincreasing⇒hasatmost1root.byobs.t=−2isasol.andalsot=−2⩽−1.✓2)0⩾t⩾−1:∣t+1∣=t+1∣t−1∣=1−t(1+t)3+(1−t)3=282(1+2t+t2+1−2t+t2−1+t2)=283t2−13=0⇒t=±133∉[−1,0]⇒nosol.in[−1,0].f(t)iseven⇒allsols.aret=±2⇒log2x=±2⇒x=2±2⇒x=14andx=4
Answered by MJS_new last updated on 27/Jan/22
x=2^t ∣t−1∣^3 +∣t+1∣^3 =28 (1) t≤−1 t^3 +3t+14=0 ⇒ t=−2 (2) −1<t≤1 t^2 −((13)/3)=0 ⇒ no solution (3) 1<t t^3 +3t−14=0 ⇒ t=2 ⇒ x=(1/4)∨x=4
x=2t∣t−1∣3+∣t+1∣3=28(1)t⩽−1t3+3t+14=0⇒t=−2(2)−1<t⩽1t2−133=0⇒nosolution(3)1<tt3+3t−14=0⇒t=2⇒x=14∨x=4

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *