Is this function $g$ also surjective? Let $f:\Bbb{R}\to\Bbb{R}$ be a surjective function. Let:
$$g(x,y)=\frac{\ln f(x) + \ln f(y)}{\ln2}-2$$
Here $g$ is restricted to the positive values of $f$. 
My question is: is the function $g(x,y)$ also surjective?
 A: Since $f$ is surjective then there exists $a$ with $f(a)=1$. Now put any $t\in\mathbb{R}$, since $f$ is surjective then there exists $s$ such that $f(s)=e^{(t+2)\ln2}$. Finally, by taking $x=s$ and $y=a$ you will have
$$
g(x,y)=\frac{\ln f(s)}{\ln 2}-2=\frac{(t+2)\ln 2}{\ln 2}-2=t
$$
as you wanted/
A: This is effectively a long comment.
The easiest way to see if a function of multiple variables is surjective (or even just to find its range) is to set all but one of the variables to a constant and see what values can be obtained as you vary the other variables. Then, for whatever values cannot be obtained in this way, see if you can somehow obtain them by varying all the variables.
In this case, the useful choice is $y_0 \in f^{-1}(1)$ (or $x_0 \in f^{-1}(1)$ as $g$ is symmetric in $x$ and $y$). The reason this is a useful choice is that $\ln f(y_0) = \ln 1 = 0$, so $$g(x, y_0) = \frac{\ln f(x)}{\ln 2} - 2.$$
Given that $f$ is surjective and $\ln$ has range $\mathbb{R}$, what values can be obtained as we vary $x$?
A: Presumably you mean surjective from R x R to R. There's a problem in the (original) definition of g because with f being surjective it takes all negative values, which can't be in the domain of g. I see this is now modified.
You would have to redefine the domain of g so that the x and y values resulted in values of f in $(0, +\infty)$ and g would then be surjective.
