[tex]\bf \stackrel{\textit{using the exponential model}}{N=2^D}~\hspace{7em}\begin{array}{ccll} \stackrel{days}{D}&\stackrel{\$}{N}\\ \cline{1-2} 1&2^1\implies 2\\ 2&2^2\implies 4 \end{array}[/tex]so, using that exponential model, the 1st output value works, but the second value of 2² does not give us 8 as output.let's check the linear model using slopes to get the equation.[tex]\bf (\stackrel{x_1}{1}~,~\stackrel{y_1}{2})\qquad (\stackrel{x_2}{2}~,~\stackrel{y_2}{8})\qquad \impliedby \begin{array}{|cc|ll} \cline{1-2} D&N\\ \cline{1-2} 1&2\\ 2&8\\ \cline{1-2} \end{array}[/tex][tex]\bf slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{8-2}{2-1}\implies \cfrac{6}{1}\implies 6 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-2=6(x-1) \\\\\\ y-2=6x-6\implies y=6x-4[/tex]now, using that model, x = 6, then y = 6(6) - 4, or y = 32.