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Given-2-log-3-x-1-log-3-x-3-8-gt-0-have-the-solution-a-x-lt-b-what-is-b-




Question Number 83774 by jagoll last updated on 06/Mar/20
Given 2(√(log_3  x−1)) − log_3  x^3  +8 > 0  have the solution a ≤ x < b.   what is b ?
Given2log3x1log3x3+8>0havethesolutionax<b.whatisb?
Answered by john santu last updated on 06/Mar/20
⇒ 2(√(log_3  x−1)) −3log_3  x + 8 >0  let (√(log_3  x−1)) = t , t ≥ 0 (i)  ⇒ 2t − 3(t^2 +1) +8 >0  3t^2  +3 −2t − 8 < 0  3t^2  −2t −5 < 0   (3t−5)(t+1) < 0 ⇒ −1 < t < (5/3) (ii)  from (i) & (ii)   0 ≤ t < (5/3) ⇒ 0 ≤ (√(log_3  x−1)) < (5/3)  0 ≤ log_3  x−1 < ((25)/9)  1 ≤ log_3  x < ((34)/9) ⇔ 3^1  ≤ x < 3^((34)/9)   so we get b = 3^((34)/9)  ⇐ the solution
2log3x13log3x+8>0letlog3x1=t,t0(i)2t3(t2+1)+8>03t2+32t8<03t22t5<0(3t5)(t+1)<01<t<53(ii)from(i)&(ii)0t<530log3x1<530log3x1<2591log3x<34931x<3349sowegetb=3349thesolution
Commented by jagoll last updated on 06/Mar/20
thank you mister
thankyoumister

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