Electric potential: Thin finite rod

Question:
A thin rod is placed along the z-axis that is stretched from $z=-d$ to $z=+d$. Let $\lambda$ be the linear density. There are two points $P_1 = (0,0,2d)$ and $P_2 = (x,0,0)$. Find the $x$ coordinate of $P_2$ so that potentials at $P_1$ and $P_2$ are equal.

Answer:
The potential at $P_1$ is positioned at the distance of $d$ from the edge of the rod, so the distance from the tiny piece of the charged rod should be $2d-z$. The charge is expressed by $\lambda dz$. Thus, the potential can be calculated as follows:
\begin{eqnarray}
\phi_1 &=& \int \frac{kdq}{r} = k\int^{d}_{-d}\frac{\lambda dz}{2d-z}  \\
  &=& -k\lambda \left[\ln(2d-z)\right]^{d}_{-d}  \\
  &=& -k\lambda (\ln(2d-d)-\ln(2d+d))  \\
  &=& -k\lambda (\ln(d)-\ln(3d))  \\
  &=& -k\lambda \left(\ln\frac{1}{3}\right)  \\
  &=& -k\lambda (-\ln(3)) \\
  &=& k\lambda \ln(3)
\end{eqnarray}
Likewise,
\begin{eqnarray}
\phi_2 &=& k\int^{d}_{-d}\frac{\lambda dz}{\sqrt{x^2+z^2}}  \\
  &=& k\lambda \ln\frac{\sqrt{x^2+d^2}+d}{\sqrt{x^2+d^2}-d}
\end{eqnarray}
These potentials are equal, so
\begin{eqnarray}
k\lambda \ln(3) &=& k\lambda \ln\frac{\sqrt{x^2+d^2}+d}{\sqrt{x^2+d^2}-d}  \\
3 &=& \frac{\sqrt{x^2+d^2}+d}{\sqrt{x^2+d^2}-d}  \\
3\sqrt{x^2+d^2}-3d &=& \sqrt{x^2+d^2}+d   \\
2\sqrt{x^2+d^2} &=& 4d  \\
x^2 + d^2 &=& 4d^2  \\
x &=& \sqrt{3}d
\end{eqnarray}

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