Answer
$$y= 5\left(\pi^{3}\right) \ln \pi(x-1)+\pi^{3}$$
Work Step by Step
Given $$y=\pi^{5 x-2}, \quad x=1$$
Since at $x=1$, $y= \pi ^3$ and
$$f'(x)= \pi^{5 x-2}\ln \pi \frac{d}{dx} (5 x-2 ) = \pi^{5 x-2}(5\ln \pi) $$
Then $ m= f'(x)\bigg|_{x=1}=5\pi^3\ln \pi $.
Hence, the tangent line is given by
\begin{align*}
\frac{y-y_1}{x- x_1}&=m\\
\frac{y-\pi^3 }{x- \pi }&=5\pi^3\ln \pi \\
y-\pi^3&= 5\pi^3\ln \pi (x-\pi )\\
y&= 5\left(\pi^{3}\right) \ln \pi(x-1)+\pi^{3}
\end{align*}
