Does every group homomorphism from $(0,\infty),\times)$ to $(\mathbb{R},+)$ send $1$ to $1$? I just have some true/false questions I am revising with and I'm not sure about this.

Let $f:((0,\infty),\times)\to(\mathbb{R},+)$ be a group homomorphism, then $f(1)=1\tag{1}$

I know that a group homomorphism is when you have 2 groups $(G,*)$ and $(H,\odot)$ such that a function $f:G \to H$ gives $f(a*b)=f(a) \odot f(b)$ for all $a,b \in G$

If someone could explain that (or give me hint), it'd be awesome!

Okay actually as I was typing this I may have had an epiphony!

I think that (1)is  false because $1\times 1 \neq 1+1$

I also have,

Let $f:((0,\infty),\times)\to(\mathbb{R},+)$ be a group homomorphism, then $f(\frac{1}{2})=-f(2)\tag{2}$

But please don't tell me the answer to this one! I'll try figure it out based on feedback from the first one and go from there
Clearly $(1)$ is not true. In general, given $f:G\rightarrow H$ group homomorphism, we know that the identity element $e_G\in G$ maps to the identity element $e_H \in H$.

You just need to identify these elements in your groups and substitute in
$$f(e_G)=e_H. \tag{1}$$

Then it should be easy to figure out $(2)$, since
$$f(1)=f\left(\frac{1}{2}\cdot 2\right) = f\left(\frac{1}{2}\right) + f(2).$$
Use the same idea to view $\frac 1 2$ and $2$. 
Try to view this 
$$f(\frac  1 2)=f(\frac 1 4.2)=2\frac 1 4\tag 1$$
but $$ -f(2)=-f(\frac 1 2.4)=4\frac 1 2\tag 2$$ which is not same.