The interval of x in which the inequality 51-4log5x2- 5 - x1-5log5x

5^1/4(log_5x)^2≥ 5 . x^1/5(log_5x) ⇒1/4(log_5x)^2 ≥ 1 + 1/5(log_5x)(log_5x) ⇒ 5(log_5x)^2 ≥ 20 + 4(log_5x)(log_5x) ⇒ (log_5x)^2 ≥ 20 log_5x ≥ 2√(5) or log_5x ...

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Byju's Answer

Standard XII

Mathematics

Fundamental Laws of Logarithms

The interval ...

Question

The interval of x in which the inequality $5^{\frac{1}{4} (l o g_{5} x)^{2}} \geq 5 \cdot x^{\frac{1}{5} (l o g_{5} x)}$

(0, 5^{- 2 \sqrt{5}}]

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[5^{2 \sqrt{5}}, \infty)

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(0, 5^{- 2 \sqrt{5}}] \cup [5^{2 \sqrt{5}}, \infty)

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(0,

\infty

)

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Solution

The correct option is C $(0, 5^{- 2 \sqrt{5}}] \cup [5^{2 \sqrt{5}}, \infty)$
$5^{\frac{1}{4} (l o g_{5} x)^{2}} \geq 5 . x^{\frac{1}{5} (l o g_{5} x)}$
$\Rightarrow \frac{1}{4} (l o g_{5} x)^{2} \geq 1 + \frac{1}{5} (l o g_{5} x) (l o g_{5} x)$
$\Rightarrow 5 (l o g_{5} x)^{2} \geq 20 + 4 (l o g_{5} x) (l o g_{5} x)$
$\Rightarrow (l o g_{5} x)^{2} \geq 20$
$l o g_{5} x \geq 2 \sqrt{5} or l o g_{5} x \leq - 2 \sqrt{5}$
but $x > 0$
So $x ϵ (0, 5^{- 2 \sqrt{5}}] \cup [5^{2 \sqrt{5}}, \infty)$

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The interval of x in which the inequality 514(log5x)2≥5⋅x15(log5x)

The correct option is C (0,5−2√5]∪[52√5,∞)514(log5x)2≥5 . x15(log5x) ⇒14(log5x)2≥1+15(log5x)(log5x) ⇒5(log5x)2≥20+4(log5x)(log5x) ⇒ (log5x)2≥20 log5x≥2√5 or log5x≤−2√5 but x>0 So x ϵ (0,5−2√5]∪[52√5,∞)

The interval of x in which the inequality $5^{\frac{1}{4} (l o g_{5} x)^{2}} \geq 5 \cdot x^{\frac{1}{5} (l o g_{5} x)}$