If log13[log4(x2−12)]<0, then x lies in the interval

log13[log4(x2−12)] 1 ⇒x2−12>4⇒x2>16 ⇒ x2−16>0⇒(x+4)(x−4)>0x∈(−∞,−4)∪(4,∞)

If log13[log4(x2−12)]<0, then x lies in the interval

log13[log4(x2−12)] 1 ⇒x2−12>4⇒x2>16 ⇒ x2−16>0⇒(x+4)(x−4)>0x∈(−∞,−4)∪(4,∞)

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If $\log_{\frac{1}{3}} [\log_{4} (x^{2} - 12)] < 0$ , then $x$ lies in the interval

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$(- \infty, - 4) \cup (4, \infty)$

$(- \infty, - 1) \cup (1, \infty)$

$(- \infty, - 5) \cup (5, \infty)$

$(- \infty, - 3) \cup (3, \infty)$

answer is D.

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Detailed Solution

$\log_{\frac{1}{3}} [\log_{4} (x^{2} - 12)] < 0$

$\log_{4} x^{2} - 12 > 1 \Rightarrow x^{2} - 12 > 4$

$\Rightarrow x^{2} > 16 \Rightarrow x^{2} - 16 > 0$

$\Rightarrow (x + 4) (x - 4) > 0$

$x \in (- \infty, - 4) \cup (4, \infty)$

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If $\log_{\frac{1}{3}} [\log_{4} (x^{2} - 12)] < 0$ , then $x$ lies in the interval