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 \sqrt{\log_{3} x - 1} - \log_{3} x^{3} + 8 > 0

h a v e the solution a ⩽ x < b .

w h a t i s b ?

Answered by john santu last updated on 06/Mar/20

\Rightarrow 2 \sqrt{\log_{3} x - 1} - {3 \log}_{3} x + 8 > 0

let \sqrt{\log_{3} x - 1} = t, t ⩾ 0 (i)

\Rightarrow 2 t - 3 (t^{2} + 1) + 8 > 0

{3 t}^{2} + 3 - 2 t - 8 < 0

{3 t}^{2} - 2 t - 5 < 0

(3 t - 5) (t + 1) < 0 \Rightarrow - 1 < t < \frac{5}{3} (ii)

from (i) & (ii)

0 ⩽ t < \frac{5}{3} \Rightarrow 0 ⩽ \sqrt{\log_{3} x - 1} < \frac{5}{3}

0 ⩽ \log_{3} x - 1 < \frac{25}{9}

1 ⩽ \log_{3} x < \frac{34}{9} \Leftrightarrow 3^{1} ⩽ x < 3^{\frac{34}{9}}

so we get b = 3^{\frac{34}{9}} \Leftarrow the solution

Commented by jagoll last updated on 06/Mar/20

thank you mister

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