If A is the set of all x - R such that x log x 2-3 log x -1-1000- and A - a - - then -10 a will be - Base of log x is 10-

A = (1000, ∞) (log x)^2 3 log x + 1 gt; log_x 10^3 = 3 log_x 10 If log_10 x=t then we have t^2 3t+1 gt; 3/t or t^3 3t^2+t 3 gt;0 or t(t^2+1) 3(t^2+ ...

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Byju's Answer

Standard IX

Mathematics

Logarithmic Inequalities

If A is the s...

Question

If A is the set of all $x ϵ R$ such that $x^{(l o g x)^{2} - 3 l o g x + 1} > 1000$ , and $A = (a, \infty)$ then $\sqrt{10 a}$ will be ___. ( (Base of logx is 10).

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Solution

$A = (1000, \infty)$
$(l o g x)^{2} - 3 l o g x + 1 > l o g_{x} 10^{3} = 3 l o g_{x} 10$
If $l o g_{10} x = t$ then we have
$t^{2} - 3 t + 1 > \frac{3}{t}$
or $t^{3} - 3 t^{2} + t - 3 > 0$
or $t (t^{2} + 1) - 3 (t^{2} + 1) > 0$
or $(t^{2} + 1) (t - 3) > 0 \Rightarrow t - 3 > 0$
as $t^{2} + 1$ is always + ive
$∴ t > 3$ or $l o g_{10} x > 3$
$∴ x > 10^{3} = 1000 ∴ x ϵ (1000, \infty)$

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If A is the set of all xϵR such that x(log x)2−3 log x+1>1000, and A=(a,∞) then √10a will be ___. ( (Base of logx is 10).

A=(1000,∞) (log x)2−3 log x+1>logx 103=3 logx 10 If log10 x=t then we have t2−3t+1>3t or t3−3t2+t−3>0 or t(t2+1)−3(t2+1)>0 or (t2+1)(t−3)>0⇒t−3>0 as t2+1 is always + ive ∴t>3 or log10x>3 ∴x>103=1000∴xϵ(1000,∞)

If A is the set of all $x ϵ R$ such that $x^{(l o g x)^{2} - 3 l o g x + 1} > 1000$ , and $A = (a, \infty)$ then $\sqrt{10 a}$ will be ___. ( (Base of logx is 10).