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Question Number 81364 by jagoll last updated on 12/Feb/20

log_2 (log_3 (log_2 (2/x))−1) < 1 has solution (1/a^(26) )<x<b. find a ?

$$\mathrm{log}_{\mathrm{2}} \:\left(\mathrm{log}_{\mathrm{3}} \:\left(\mathrm{log}_{\mathrm{2}} \:\frac{\mathrm{2}}{\mathrm{x}}\right)−\mathrm{1}\right)\:<\:\mathrm{1}\: \\ $$ $$\mathrm{has}\:\mathrm{solution}\:\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{26}} }<\mathrm{x}<\mathrm{b}.\:\mathrm{find}\:\mathrm{a}\:? \\ $$

Commented byjohn santu last updated on 12/Feb/20

(i) log_3 (log_2 (2/x))−1>0 (ii) log_3 (log_2 (2/x))−1 <2 ⇒ 0 < log_3 (log_2 (2/x))−1 <2 ⇒ 1 < log_3 (log_2 (2/x)) < 3 ⇒ 3 < log_2 ((2/x)) < 27 ⇒ 2^3 <(2/x) < 2^(27) ⇒ (1/2^(27) ) < (x/2) < (1/2^3 ) ⇒ (1/2^(26) ) < x < (1/2^2 ) . we get a = 2

$$\left(\mathrm{i}\right)\:\mathrm{log}_{\mathrm{3}} \:\left(\mathrm{log}_{\mathrm{2}} \:\frac{\mathrm{2}}{\mathrm{x}}\right)−\mathrm{1}>\mathrm{0} \\ $$ $$\left(\mathrm{ii}\right)\:\mathrm{log}_{\mathrm{3}} \:\left(\mathrm{log}_{\mathrm{2}} \:\frac{\mathrm{2}}{\mathrm{x}}\right)−\mathrm{1}\:<\mathrm{2} \\ $$ $$\Rightarrow\:\mathrm{0}\:<\:\mathrm{log}_{\mathrm{3}} \:\left(\mathrm{log}_{\mathrm{2}} \:\frac{\mathrm{2}}{\mathrm{x}}\right)−\mathrm{1}\:<\mathrm{2} \\ $$ $$\Rightarrow\:\mathrm{1}\:<\:\mathrm{log}_{\mathrm{3}} \:\left(\mathrm{log}_{\mathrm{2}} \:\frac{\mathrm{2}}{\mathrm{x}}\right)\:<\:\mathrm{3} \\ $$ $$\Rightarrow\:\mathrm{3}\:<\:\mathrm{log}_{\mathrm{2}} \:\left(\frac{\mathrm{2}}{\mathrm{x}}\right)\:<\:\mathrm{27}\: \\ $$ $$\Rightarrow\:\mathrm{2}^{\mathrm{3}} \:<\frac{\mathrm{2}}{\mathrm{x}}\:<\:\mathrm{2}^{\mathrm{27}} \: \\ $$ $$\Rightarrow\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{27}} }\:<\:\frac{\mathrm{x}}{\mathrm{2}}\:<\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} }\: \\ $$ $$\Rightarrow\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{26}} }\:<\:\mathrm{x}\:<\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }\:.\:\mathrm{we}\:\mathrm{get}\:\mathrm{a}\:=\:\mathrm{2}\: \\ $$

Commented byjagoll last updated on 12/Feb/20

thank mister

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

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