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Given data:

$\begin{array}{|ll|}\hline \text{year since}1700& \text{population (in millions)}\\ 0& 2\\ 50& 2\\ 100& 7\\ 150& 26\\ 200& 172\\ 250& 307\\ 299& 337\\ 308& 345\\ 310& 351\\ 312& 32\\ \hline\end{array}$

Log of given data:

$\begin{array}{|lll|}\hline \text{year since}1700& \text{population (in millions)}& \log \text{(population)}\\ 0& 2& 0.301029996\\ 50& 2& 0.301029996\\ 100& 7& 0.84509804\\ 150& 26& 1.414973348\\ 200& 82& 1.913813852\\ 250& 307& 2.235528447\\ 299& 337& 2.487138375\\ 308& 345& 2.527629901\\ 310& 351& 2.537819095\\ 312& 2.545307116& \\ \hline\end{array}$

Making use of a $Ti83/84$ calculator

Step 1: Select STAT;

Step 2: Select 1: EDIT

Step 3: Type the data for each year since $1700$in list L1 and the population logarithmic in list L2.

Step 4: Hit STAT once more, select CALC, and then $LinReg(a+bx).$

The required result:

$y=a+bx$

$a=0.1565$

$b=0.0079$

Substituting the value in $a$and $b$:

$y=0.1565+0.0079x$

Since $1700$, $x$denotes the year, while $y$is the population logarithm.

Given data:

$\begin{array}{|ll|}\hline \text{year since}1700& \text{population (in millions)}\\ 0& 2\\ 50& 2\\ 100& 7\\ 150& 26\\ 200& 82\\ 250& 172\\ 299& 307\\ 308& 337\\ 310& 351\\ 312& 25\\ \hline\end{array}$

Substituting the value $x$by $320$:

$\log y=0.1565+0.0079x$

$\log y=0.1565+0.0079\left(320\right)$

$\log y=2.6845$

Using the exponential function with a base of ten

$y={10}^{12.6845}$

$={10}^{2.6845}$

$=483.615$