Proving that $V$ is a vector space, if $a+b=ab$ and $a *b=a^b$. I am currently studying Halmos' "Linear Algebra Problem Book" and am stuck on problem 21(4). 
Let $V$ be the set $\mathbb{R}_+$, and let $F$ be the set $\mathbb{R}$.
Let's define the sum of two positive numbers: 
$$a+b = ab,$$
and the product of a positive number $a$ and a real number $b$: 
$$a*b = b^a.$$
Prove that $V$ is a vector-space.
Proving that addition is commutative, associative, as well as the existence of an additive identity (which is equal to $1$) and an inverse (which is equal to $1/x$) wasn't problematic.
However, multiplication and distributivity caused a few problems. Is the multiplicative identity also $1$, since $x^1 = x$?
Is $(ab) x = x^{b^a}$, in which case isn't $a (bx) = {x^b}^a = x^{ba}$?

Yes, the scalar $1 \in \Bbb R$ is the multiplicative identity for the vector space, because $x^1=x$, as you wrote.

Actually, $(ab)x \neq x^{a^b}$ because since $a,b$ are scalars, their product is the usual multiplication on $\Bbb R$, not the scalar multiplication as defined for a scalar and a vector.
Hence, $(ab)x=x^{ab}=x^{ba}=(x^b)^a=a(x^b)=a(bx)$, so scalar multiplication is associative.

For distributivity, what you need is $(a+b)x=ax+bx$ and $a(x+y)=ax+yx$.
The first one: $(a+b)x=x^{a+b}=x^ax^b=ax+bx$
The second one: $a(x+y)=a(xy)=(xy)^a=x^ay^a=ax+ay$.
