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Hello
$\displaystyle \begin{aligned}
\ln N&=\ln\left(N_0 \mathrm{e}^{-\lambda t}\right)\\
&=\ln N_0 + \ln \left( \mathrm{e}^{-\lambda t}\right)\\
&=\ln N_0-\lambda t
\end{aligned}$
so the equation of the straight line is $\displaystyle y=\ln N_0 -\lambda t$. Acknowledged that the gradient is the coefficient which multiplies $\displaystyle t$ and that the intercept is $\displaystyle y(0)$, can you find these two constants ?
$\displaystyle N = N_0 e^{- \lambda t}$
$\displaystyle \Rightarrow \ln N = \ln (N_0 e^{- \lambda t})$
$\displaystyle \Rightarrow \ln N = \ln N_0 + \ln (e^{- \lambda t})$
$\displaystyle \Rightarrow \ln N = - \lambda t + \ln N_0$
now remember the slope-intercept form of the equation of a line. i leave the rest to you