## Calculus (3rd Edition)

$$y= 5\left(\pi^{3}\right) \ln \pi(x-1)+\pi^{3}$$
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*}