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Observe the function

$f\left(t\right)=c_0+c_{1}t$

Where$f\left(t\right)=\log \left(D\right)$

Assume that t=0in 1975.

Then,$f\left(0\right)=\log \left(533\right),f\left(10\right)=\log \left(1823\right),f\left(20\right)=\log \left(4974\right),f\left(30\right)=\log \left(7933\right)$

Then, we have a linear system that needs to solve.

$\left[\begin{array}{l}f\left(t_1\right)=\log \left(D_1\right)\\ f\left(t_2\right)=\log \left(D_2\right)\\ f\left(t_3\right)=\log \left(D_3\right)\\ f\left(t_4\right)=\log \left(D_4\right)\end{array}\right]or\left[\begin{array}{l}{c}_0=\log \left(533\right)\\ c_0+10c_1=\log \left(1823\right)\\ c_0+20c_1=\log \left(4974\right)\\ c_0+30c_1=\log \left(7933\right)\end{array}\right]$

$A_{c_1}^{c_0}=\stackrel{鈫�}{b}$where,

$A=\left[\begin{array}{l}10\\ 110\\ 120\\ 130\end{array}\right],\stackrel{鈫�}{b}=\left[\begin{array}{l}\log \left(533\right)\\ \log \left(1823\right)\\ log\left(4974\right)\\ \log \left(7933\right)\end{array}\right]$

Now, we are going to find the least solution using the formulas below.

$\begin{array}{l}{c}_0=\frac{\left({}_{i}t_i^2\right)\left({}_{i}b_i\right)鈭�\left({}_{i}t_i\right)\left({}_{i}t_i{b}_i\right)}{n\left({}_{i}t_i^2\right)鈭�{\left({}_{i}t_i\right)}^2}\\ =\frac{\left(1400\right)\left(13.584\right)鈭�\left(60\right)\left(223.52\right)}{4\left(1400\right)鈭�{\left(60\right)}^2}\\ 鈮�2.8032\\ c_1=\frac{n\left({}_{i}t_i{b}_i\right)鈭�\left({}_{i}b_i\right)\left({}_{i}t_i\right)}{n\left({}_{i}t_i^2\right)鈭�{\left({}_{i}t_i\right)}^2}\\ =\frac{4\left(223.52\right)鈭�\left(13.584\right)鈭�\left(60\right)}{4\left(1400\right)鈭�{\left(60\right)}^2}\\ 鈮�0.0395\\ 鈭�\log \left(D\right)=2.8032+0.0395t\end{array}$

Convert the logarithm function to an exponent equation we get

$\begin{array}{l}\log \left(D\right)=2.8032+0.0395t\\ 鈬�D={10}^{2.8032+0.0395t}\\ 鈬�D\left(t\right)=\left(635.33\right){10}^{0.0395t}\end{array}$

When (t=40) the debt that the formula from part (a) will predict is

$\begin{array}{l}D\left(40\right)=\left(635.33\right){10}^{0.0395\left(40\right)}\\ =24,154.57\end{array}$

Hence,

The solution of part (a) is $\log \left(D\right)=2.803+0.0395t,D\left(t\right)=\left(635.33\right){10}^{0.0395t}$

The solution of part (b) is $D\left(40\right)=24,154.57$.