Answer
The exact length of the polar curve:
$$ r=5^{\theta} \quad \text {for } \theta =0 \quad \text {to} \quad \theta=2 \pi $$
is equal to
$$
\begin{aligned} L &=\int_{a}^{b} \sqrt{r^{2}+(d r / d \theta)^{2}} d \theta \\
&=\frac{\sqrt{1+(\ln 5)^{2}}}{\ln 5}\left(5^{2 \pi}-1\right) \end{aligned}
$$
Work Step by Step
The exact length of the polar curve:
$$ r=5^{\theta} \quad \text {for } \theta =0 \quad \text {to} \quad \theta=2 \pi $$
is equal to
$$
\begin{aligned} L &=\int_{a}^{b} \sqrt{r^{2}+(d r / d \theta)^{2}} d \theta \\
& =\int_{0}^{2 \pi} \sqrt{\left(5^{\theta}\right)^{2}+\left(5^{\theta} \ln 5\right)^{2}} d \theta \\
& =\int_{0}^{2 \pi} \sqrt{5^{2 \theta}\left[1+(\ln 5)^{2}\right]} d \theta \\ &=\sqrt{1+(\ln 5)^{2}} \int_{0}^{2 \pi} \sqrt{5^{2 \theta}} d \theta \\
&=\sqrt{1+(\ln 5)^{2}} \int_{0}^{2 \pi} 5^{\theta} d \theta \\
& =\sqrt{1+(\ln 5)^{2}}\left[\frac{5^{\theta}}{\ln 5}\right]_{0}^{2 \pi} \\ &=\sqrt{1+(\ln 5)^{2}}\left(\frac{5^{2 \pi}}{\ln 5}-\frac{1}{\ln 5}\right) \\ &=\frac{\sqrt{1+(\ln 5)^{2}}}{\ln 5}\left(5^{2 \pi}-1\right) \end{aligned}
$$
