# 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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