GAPlus(1,R) is rationally powered

Statement
The group $$GA^+(1,\R)$$, denoted GAPlus(1,R), defined explicitly as the group under composition of maps from $$\R$$ to $$\R$$ of the form:

$$x \mapsto ax + b, \qquad a,b \in \R, a > 0$$

is a rationally powered group. In other words, for any positive integer $$n$$ and any element $$u \in GA^+(1,\R)$$, there is a unique $$v \in GA^+(1,\R)$$ such that $$v^n = u$$.

Proof
Suppose $$u$$ is the map:

$$x \mapsto a_1x + b_1$$

We want to find all elements $$v$$ that are maps of the form $$x \mapsto a_2x +b_2$$ such that $$v^n = u$$. By composing $$v$$ with itself $$n$$ times, we get

$$v^n = x \mapsto a_2^nx + b_2(1 + a_2 + \dots + a_2^{n-1})$$

For this to equal $$u$$, we know that the coefficient of $$x$$ and the constant term should match up separately, so we get:

$$a_1 = a_2^n$$

and:

$$b_1 = b_2(1 + a_2 + \dots + a_2^{n-1})$$

Solving, we get that the unique solution is the element $$v$$ with:

$$a_2 = a_1^{1/n}, \qquad b_2 = \frac{b_1}{1 + a_1^{1/n} + \dots + a_1^{(n-1)/n}}$$

In other words, the solution is:

$$x \mapsto a_1^{1/n}x + \frac{b_1}{1 + a_1^{1/n} + \dots + a_1^{(n-1)/n}}$$