The range of values of $^{'} x^{'}$ that satisfies the inequation ${log}_{2} {log}_{(0.5)} (2^{x} \frac{15}{16}) \leq 2$ is

x \in [{log}_{2} \frac{1}{15}, {log}_{2} \frac{16}{15})

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x \in ({log}_{2} \frac{1}{15}, {log}_{2} \frac{16}{15})

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x \in [{log}_{2} \frac{1}{15}, {log}_{2} \frac{16}{15}]

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x \in ({log}_{2} \frac{1}{15}, {log}_{2} \frac{16}{15}]

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Solution

The correct option is B $x \in [{log}_{2} \frac{1}{15}, {log}_{2} \frac{16}{15})$
Given,
$0 < {log}_{2} l o g_{(0.5)} (2^{x} \frac{15}{16}) \leq 2$

$= 0 < {log}_{0.5} (2^{x} \frac{15}{16}) \leq 4$

$= (∵ l o g_{a} (b) = x \Rightarrow b = a^{x})$

$= 0 > {log}_{2} 2^{x} (\frac{15}{16}) \geq - 4$

$= 1 > 2^{x} (\frac{15}{16}) \geq \frac{1}{16}$

$= \frac{16}{15} > 2^{x} \geq \frac{1}{15}$

$t a k i n g l o g, w e g e t$
$x \in [{log}_{2} \frac{1}{15}, {log}_{2} \frac{16}{15})$

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Similar questions

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x

{log}_{2} {log}_{0.5} (2^{x} \frac{15}{16}) \leq 2

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