Question

Range of the function $f\left(x\right)=\sqrt{{\log }_{\frac{1}{5}}\left(\frac{{16}^x}{{16.2}^{x^2}}\right)-2}$ is

• A. $\varphi$
• B. $[0,\infty )$
• C. $(0,1]$
• D. $(0,1)\cup (1,\infty )$

Answer 1

The correct option is B $[0,\infty )$

$f\left(x\right)=\sqrt{lo{g}_{1/5}\left(\frac{{16}^x}{{16.2}^{x^2}}\right)-2}$

$=\sqrt{\frac{ln\left(\frac{{16}^x}{{16.2}^{x^2}}\right)}{ln\left(1/5\right)}-2}$

$=\sqrt{\frac{ln\left(\frac{{16}^{x-1}}{2^{x^2}}\right)}{-ln5}-2}$

$=\sqrt{-\frac{ln\left(2^{4x-4-x^2}\right)}{ln5}-2}$

$=\sqrt{-\frac{lo{g}_2\left(2^{(4x-4-x^2)}\right)}{ln5lo{g}_{2}e}-2}$

$=\sqrt{-\frac{(4x-4-x^2)}{\frac{lo{g}_{2}5}{lo{g}_{2}e}lo{g}_{2}e}-2}$

$=\sqrt{\frac{x^2-4x+4}{lo{g}_{2}5}-2}$

$=\sqrt{\frac{(x-2{)}^2}{lo{g}_{2}5}-2}$

Now

$-\infty <x<\infty$

$-\infty <x-2<\infty$

$0<(x-2{)}^2<\infty$

$0<\frac{(x-2{)}^2}{lo{g}_{2}5}<\infty$

$-2<\frac{(x-2{)}^2}{lo{g}_{2}5}-2<\infty$

$0\le \sqrt{\frac{(x-2{)}^2}{lo{g}_{2}5}-2}<\infty$

$\Rightarrow$ Range $\left(f\right)$ = $[0,\infty )$