Find the range of the function \(log_{\sqrt{2}}(2-log_2(16sin^2x+1))\)

Solution :

Now 1 \(\le\) \(16sin^2x\) + 1) \(\le\) 17

0 \(\le\) \(log_2(16sin^2x+1)\) \(\le\) \(log_217\)

2 – \(log_217\) \(\le\) 2 – \(log_2(16sin^2x+1)\) \(\le\) 2

Now consider 0 < 2 – \(log_2(16sin^2x+1)\) \(\le\) 2

-\(\infty\) < \(log_{\sqrt{2}}(2-log_2(16sin^2x+1))\) \(\le\) \(log_{\sqrt{2}}2\) = 2

the range is (-\(\infty\), 2]

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