For t = 2:
[tex]\begin{gathered} k(2)=2(2)-1 \\ k(2)=4-1 \\ k(2)=3 \end{gathered}[/tex]If k(t) = 7:
[tex]\begin{gathered} k(t)=7=2t-1 \\ solve_{\text{ }}for_{\text{ }}t\colon_{} \\ 7=2t-1 \\ 7+1=2t \\ 2t=8 \\ t=\frac{8}{2} \\ t=4 \end{gathered}[/tex]