= Solution
Write $M=M_{\rm Pl}$ for the <reduced Planck mass>. For a nontrivial power law with $\lambda\ne0$, $H=p/t$, $\dot\varphi=BM/t$, and $V=V_0(t/t_0)^{-\lambda B}$. Matching its time dependence to the kinetic and <Friedmann equation> terms requires $\lambda B=2$. The <scalar field equation of motion> and the <Friedmann equation> then reduce to
$$
B(3p-1)M^2=\lambda V_0t_0^2,\qquad
3M^2p^2=\frac{B^2M^2}{2}+V_0t_0^2.
$$
Solving gives the <exponential-potential scalar-dominated power-law solution>
$$
\boxed{p=\frac2{\lambda^2},\qquad B=\frac2\lambda,\qquad
t_0^2=\frac{2M^2(6-\lambda^2)}{\lambda^4V_0}.}
$$
Choose $t>0$ and the positive square root for $t_0$. The <existence conditions for exponential-potential scaling> matter: if $V_0>0$, a finite real $t_0$ requires $0<\lambda^2<6$, rather than an inclusive bound on signed $\lambda$. At $\lambda^2=6$, nonzero $V_0$ is incompatible with this logarithmic ansatz. If $V_0<0$, the corresponding scaling solution instead requires $\lambda^2>6$. The sign of $\lambda$ changes the field's rolling direction, not $p$. For $\lambda=0$ the potential is constant and this finite-$p$ ansatz does not apply; with positive $V_0$, a constant field gives <de Sitter spacetime> expansion.
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