Why does $v(x_1, x_2)=x_1^cx_2^d$ with $a=\frac{c}{c+d}$ imply $v(x_1, x_2)=x_1^ax_2^{1-a}$?

Question

If we assume that $v(x_1, x_2)=x_1^cx_2^d$.

If we set $a:=\frac{c}{c+d}$. Why does this imply that $v(x_1, x_2)=x_1^ax_2^{1-a}$?

I don't see how $c^{-1} = 1-a$.

That would mean that $\frac{1}{c}+\frac{c}{c+d}=c^{-1}+d^{-1}=1 \implies c+d=1$.

But this is only true iff $c+d =1$, yet there's nothing that tells me anything about this assumption.

Is there anything I missed?

I expect both $c$ and $d$ to be within $\mathbb{R}^+ \backslash \{0\}$.