The number of integers satisfying the inequal | vedclass

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(B) Let $t = \sqrt{\log_3(x) - 1}$. For the square root to be defined,$\log_3(x) - 1 \ge 0$,so $\log_3(x) \ge 1$,which implies $x \ge 3$. Also,$t \ge

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$6$

Vedclass · Answer

(B) Let $t = \sqrt{\log_3(x) - 1}$. For the square root to be defined,$\log_3(x) - 1 \ge 0$,so $\log_3(x) \ge 1$,which implies $x \ge 3$. Also,$t \ge 0$.Then $\log_3(x) = t^2 + 1$.The inequality becomes $t + \frac{\frac{1}{2} \cdot 3 \log_3(x)}{-1} + 2 > 0$.Substituting $\log_3(x) = t^2 + 1$,we get $t - \frac{3}{2}(t^2 + 1) + 2 > 0$.$t - \frac{3}{2}t^2 - \frac{3}{2} + 2 > 0 \Rightarrow -\frac{3}{2}t^2 + t + \frac{1}{2} > 0$.Multiplying by $-2$,we get $3t^2 - 2t - 1 Factoring the quadratic: $(3t + 1)(t - 1) This holds for $-\frac{1}{3} Substituting back $t = \sqrt{\log_3(x) - 1}$,we get $0 \le \sqrt{\log_3(x) - 1} Squaring gives $0 \le \log_3(x) - 1 This implies $3^1 \le x The integers satisfying this are $3, 4, 5, 6, 7, 8$.There are $6$ such integers.

The number of integers satisfying the inequality $\sqrt {{{\log }_3}(x) - 1} + \frac{{\frac{1}{2}{{\log }_3}({x^3})}}{{{{\log }_3}(\frac{1}{3})}} + 2 > 0$ is

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