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For $V_0\ne 0$, real nonzero $\lambda$, and a finite positive logarithmic time scale $t_0$, the exponential-potential scalar-dominated power-law solution requires the displayed strict sign condition. Thus positive $V_0$ permits $0<{\lambda }^2<6$, while negative $V_0$ permits ${\lambda }^2>6$. The boundary ${\lambda }^2=6$ cannot occur with nonzero $V_0$ in this ansatz. Setting $\lambda =0$ gives a constant potential and a separate branch, including constant-field de Sitter spacetime for $V_0>0$.

Write $M=M_{\mathrm{Pl}}$ for the reduced Planck mass. For a nontrivial power law with $\lambda \ne 0$, $H=p/t$, $\dot{\phi }=BM/t$, 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 toSolving gives the exponential-potential scalar-dominated power-law solutionChoose $t>0$ 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.