The table gives the midyear population of Japan, in thousands, from 1960 to 2010.

Step 1
Following the suggestions given by the problem, let
$t=$ the number of years after 1960 (so $t=0$ is 1960)
enter the population numbers with 94000 subtracted from each.
Using Desmos, first add a table. (clic on "$+$" at the upper left) and enter the numbers. It should look something like in this image:
Step 2
$\begin{array}{|cc|}\hline x_1& y_1\\ 0& 94092-94000\\ 5& 98883-94000\\ 10& 104345-94000\\ 15& 111573-94000\\ 20& 116807-94000\\ 25& 120754-94000\\ 30& 123537-94000\\ 35& 125327-94000\\ 40& 126776-94000\\ 45& 127715-94000\\ 50& 127579-94000\\ \hline\end{array}$
$y_1\sim a{b}^{x_1}$
$R^2=0.7692e_1$
$a=10665.3b=1.02682$
Step 3
Next, you can get an exponential form $y=a{b}^x$ by typing in the next box below the one with the table:
$y_1\to \sim a{b}^{x_1}$
The values for a and b appear below. It should look similar to what is the lower right in the image above. Desmos gives a different answer than the book's. (Compare it in the graph at the bottom). The model with the proper variable names and shifted back up by 94000 would be:
$P(t)=10665.3(1.02682{)}^t+94000$
Step 4
Next in another box below, type this to get a logistic model:
$y_1\to \sim /M\downarrow 1+A{e}^{-k{x}_1}$
$y_1\sim a{b}^{x_1}$
$P{R}^2=0.7692e_1$
$a=10665.3b=1.02682$
$y_1\sim \frac{M}{1+A{e}^{-k{x}_1}}$
$R^2=0.9927e_2$
$M=33086.4A=12.3428$
$k=0.165732$
Step 5
This time it does given an answer similar to the book's. Using the right variables and shifting back up by 94000, we have:
$P(t)=\frac{33086.4}{1+12.3428e^{-0.165732t}}+94000$
Step 6
Using GeoGebra 5 and entering the same numbers in as a list of points, the command FitExp does give the same answer as the book. (shifted it back up by 94000 here)
$P(t)=1909.7761e^{0.07655}t+94000$
$=1909.7761(e^{{0.07655}^t})+94000$
$=1909.7761(1.0796{)}^t+94000$
Step 7
The logistic model is better than the exponential. The Desmos one seems to stay closer to the points overall than the other exponential one.