Answer
$L=\frac{\sqrt{1+(\ln(5))^{2}}}{\ln(5)}\left[5^{2\pi}-1\right]$
Work Step by Step
The required length is:
$$L=\int_{0}^{2\pi}\sqrt{(5^{\theta})^{2}+(\ln(5)5^{\theta})^{2}}d\theta$$
$$L=\sqrt{1+(\ln(5))^{2}}\int_{0}^{2\pi}\sqrt{(5^{\theta})^{2}}d\theta$$
$$L=\sqrt{1+(\ln(5))^{2}}\int_{0}^{2\pi}5^{\theta}d\theta$$
$$L=\sqrt{1+(\ln(5))^{2}}\left[\frac{1}{\ln(5)}5^{\theta}\right]_{0}^{2\pi}$$
$$L=\sqrt{1+(\ln(5))^{2}}\left[\frac{1}{\ln(5)}5^{2\pi}-\frac{1}{\ln(5)}5^{0}\right]$$
$$L=\sqrt{1+(\ln(5))^{2}}\left[\frac{1}{\ln(5)}5^{2\pi}-\frac{1}{\ln(5)}\right]$$
$$L=\frac{\sqrt{1+(\ln(5))^{2}}}{\ln(5)}\left[5^{2\pi}-1\right]$$
