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log-2-log-3-log-2-2-x-1-lt-1-has-solution-1-a-26-lt-x-lt-b-find-a-




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 by john 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 by jagoll last updated on 12/Feb/20
thank mister
$$\mathrm{thank}\:\mathrm{mister} \\ $$

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