Question

Range of the function $f\left(x\right)={\log }_{\sqrt{2}}(2-{\log }_2(16{\sin }^{2}x+1))$ is:

Answer 1

$0<{\sin }^{2}x<1$

$0<16{\sin }^{2}x<16$

$1<16{\sin }^{2}x+1<17$

$lo{g}_{2}1<lo{g}_2(16{\sin }^{2}x+1)<lo{g}_{2}17$

$-lo{g}_{2}1>-lo{g}_2(16{\sin }^{2}x+1)>-lo{g}_{2}17$

$2-lo{g}_{2}1>2-lo{g}_2(16{\sin }^{2}x+1)>2-lo{g}_{2}17$

$lo{g}_{\sqrt{2}}(2-lo{g}_{2}1)>lo{g}_{\sqrt{2}}(2-lo{g}_2(16{\sin }^{2}x+1\left)\right)>lo{g}_{\sqrt{2}}(2-lo{g}_{2}17)$.