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Given data:

$\begin{array}{|ll|}\hline \text{year since}1700& \text{population (in millions)}\\ 0& 125\\ 50& 163\\ 100& 203\\ 150& 276\\ 200& 408\\ 250& 729\\ 299& 732\\ 308& 740\\ 310& 729\\ 312& 73\\ \hline\end{array}$

Log of given data:

$\begin{array}{|lll|}\hline \text{year since}1700& \text{population (in millions)}& \log \text{(population)}\\ 0& 125& 2.096910013\\ 50& 163& 2.212187604\\ 100& 203& 2.307496038\\ 150& 276& 2.440909082\\ 200& 408& 2.610660163\\ 250& 547& 2.737987326\\ 299& 729& 2.862727528\\ 308& 732& 2.864511081\\ 310& 738& 2.868056362\\ 312& 740& 2.86923172\\ \hline\end{array}$

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:

$y=a+bx$

$a=2.0796$

$b=0.0026$

Substituting the value in $a$and $b$

$y=2.0796+0.0026x$

The year since $1700$is represented by $x$, while the population logarithm is represented by $y$.

Given data:

$\begin{array}{|ll|}\hline \text{year since}1700& \text{population (in millions)}\\ 0& 125\\ 50& 163\\ 100& 203\\ 150& 276\\ 200& 408\\ 250& 729\\ 299& 732\\ 308& 740\\ 310& 729\\ 312& 73\\ \hline\end{array}$

Substituting the value $x$by $320$:

$\log y=2.0796+0.0026x$

$\log y=2.0796+0.0026\left(320\right)$

$\log y=2.9116$

Using the exponential function with a base of ten

$y={10}^{\log y}$

$={10}^{2.9116}$

$=815.831$