The amount of money in an account may increase due to rising stock prices and decrease due to falling stock prices. Nathan is studying the change in the amount of money in two accounts, A and B, over time.
The amount f(x), in dollars, in account A after x years is represented by the function below:
f(x) = 10,125(1.83)x
Part A: The table below shows the amount g(r), in dollars, of money in account B after r years.
r (number of years)
1
2
3
4
g(r) (amount in dollars)
9,638
18,794.10
36,648.50
71,464.58
Which account recorded a greater percentage change in amount of money over the previous year? Justify your answer.

It looks like your first account increases linearly,
unless you have a typo and meant:
f(x) = 10,125(1.83)^x
the second is
exponential and can be written as
g(r) = 9,638(1.95)^r
clearly g(r) has a higher percentage change since 1.95 > 1.83
a more interesting question might have been: When will the amount be the same?
10,125(1.83)^x = 9,638(1.95)^x
10,125(1.83)^x = 9,638(1.95)^x
(1.83)^x = .951901...(1.95)^x
take log of both sides and use log rules
xlog1.83 = log .951901... + xlog 1.95
x(log 1.83  log 1.95) = log .951901...
x = .776..
check:
f(.776..) = 10,125(1.83)^.776.. = 1618410
g(.776..) = 9,638(1.95)^.776... = 16184.10

Thank you so much!!! :)

How did you get the 1.95 in the second equation


unless you have a typo and meant:
f(x) = 10,125(1.83)^x
the second is
exponential and can be written as
g(r) = 9,638(1.95)^r
clearly g(r) has a higher percentage change since 1.95 > 1.83
a more interesting question might have been: When will the amount be the same?
10,125(1.83)^x = 9,638(1.95)^x
10,125(1.83)^x = 9,638(1.95)^x
(1.83)^x = .951901...(1.95)^x
take log of both sides and use log rules
xlog1.83 = log .951901... + xlog 1.95
x(log 1.83  log 1.95) = log .951901...
x = .776..
check:
f(.776..) = 10,125(1.83)^.776.. = 1618410
g(.776..) = 9,638(1.95)^.776... = 16184.10