Finding the domain and the range of this function Problem: Find the domain and range of the function $f(x)=\log \sqrt{4-x^2}$
My attempt: 
For domain
$y=f(x)=\log \sqrt{4-x^2}$
$\implies y=\log \sqrt{4-x^2}$
Since
logarithm of only positive numbers is possible,
$4-x^2>0$
$\implies x^2-4<0$
$\implies(x+2)(x-2)<0$
$\implies -2<x<2$
Therefore $D_f=(-2,2)$
For range 
$y=f(x)=\log \sqrt{4-x^2}$
$y=\log \sqrt{4-x^2}$
$y=\frac{1}{2}\log (4-x^2)$
$2y=\log (4-x^2)$
$e^{2y}=4-x^2$
$x=\sqrt{4-e^{2y}}$
Since the exponent can take any real value, $R_f=(-\infty,+\infty)$
My problem: The answer given in my book for $D_f$ matches with my $D_f$ but $R_f=(-\infty,\log 2]$ according to my book. Where have i gone wrong?

$\sqrt {4-x^2}$ cannot be greater than $\sqrt 4=2$, so $\log \sqrt{4-x^2}$ cannot be any greater than $\log \sqrt 4=\log 2$.  It achieves that value when $x=0.$  As $\sqrt{4-x^2}$ can achieve $0$, you can have $\log \sqrt{4-x^2}$ as negative as you want, so the range is $(-\infty, \log 2]$
