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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 ?

$$\mathrm{Given}\:\mathrm{2}\sqrt{\mathrm{log}_{\mathrm{3}} \:{x}−\mathrm{1}}\:−\:\mathrm{log}_{\mathrm{3}} \:{x}^{\mathrm{3}} \:+\mathrm{8}\:>\:\mathrm{0} \\ $$ $${have}\:\mathrm{the}\:\mathrm{solution}\:\mathrm{a}\:\leqslant\:{x}\:<\:{b}.\: \\ $$ $${what}\:{is}\:{b}\:?\: \\ $$

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

$$\Rightarrow\:\mathrm{2}\sqrt{\mathrm{log}_{\mathrm{3}} \:{x}−\mathrm{1}}\:−\mathrm{3log}_{\mathrm{3}} \:{x}\:+\:\mathrm{8}\:>\mathrm{0} \\ $$ $$\mathrm{let}\:\sqrt{\mathrm{log}_{\mathrm{3}} \:{x}−\mathrm{1}}\:=\:\mathrm{t}\:,\:\mathrm{t}\:\geqslant\:\mathrm{0}\:\left(\mathrm{i}\right) \\ $$ $$\Rightarrow\:\mathrm{2t}\:−\:\mathrm{3}\left(\mathrm{t}^{\mathrm{2}} +\mathrm{1}\right)\:+\mathrm{8}\:>\mathrm{0} \\ $$ $$\mathrm{3t}^{\mathrm{2}} \:+\mathrm{3}\:−\mathrm{2t}\:−\:\mathrm{8}\:<\:\mathrm{0} \\ $$ $$\mathrm{3t}^{\mathrm{2}} \:−\mathrm{2t}\:−\mathrm{5}\:<\:\mathrm{0}\: \\ $$ $$\left(\mathrm{3t}−\mathrm{5}\right)\left(\mathrm{t}+\mathrm{1}\right)\:<\:\mathrm{0}\:\Rightarrow\:−\mathrm{1}\:<\:\mathrm{t}\:<\:\frac{\mathrm{5}}{\mathrm{3}}\:\left(\mathrm{ii}\right) \\ $$ $$\mathrm{from}\:\left(\mathrm{i}\right)\:\&\:\left(\mathrm{ii}\right)\: \\ $$ $$\mathrm{0}\:\leqslant\:\mathrm{t}\:<\:\frac{\mathrm{5}}{\mathrm{3}}\:\Rightarrow\:\mathrm{0}\:\leqslant\:\sqrt{\mathrm{log}_{\mathrm{3}} \:{x}−\mathrm{1}}\:<\:\frac{\mathrm{5}}{\mathrm{3}} \\ $$ $$\mathrm{0}\:\leqslant\:\mathrm{log}_{\mathrm{3}} \:{x}−\mathrm{1}\:<\:\frac{\mathrm{25}}{\mathrm{9}} \\ $$ $$\mathrm{1}\:\leqslant\:\mathrm{log}_{\mathrm{3}} \:{x}\:<\:\frac{\mathrm{34}}{\mathrm{9}}\:\Leftrightarrow\:\mathrm{3}^{\mathrm{1}} \:\leqslant\:{x}\:<\:\mathrm{3}^{\frac{\mathrm{34}}{\mathrm{9}}} \\ $$ $$\mathrm{so}\:\mathrm{we}\:\mathrm{get}\:\mathrm{b}\:=\:\mathrm{3}^{\frac{\mathrm{34}}{\mathrm{9}}} \:\Leftarrow\:\mathrm{the}\:\mathrm{solution} \\ $$ $$ \\ $$

Commented byjagoll last updated on 06/Mar/20

thank you mister

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{mister} \\ $$

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