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.
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
The amount of money in an account may increase due to rising stock prices and decrease due to falling stock prices. Marco 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
To determine which account recorded a greater percentage change in the amount of money over the previous year, we need to compare the percentage change for each account.
For account A, we can calculate the amount of money after each year by plugging in the values of x from the table into the function f(x) = 10,125(1.83)^x:
f(1) = 10,125(1.83)^1 = 18,529.75
f(2) = 10,125(1.83)^2 = 34,023.73
f(3) = 10,125(1.83)^3 = 62,366.94
f(4) = 10,125(1.83)^4 = 114,211.25
To calculate the percentage change in account A over the previous year, we subtract the value of the previous year from the current year and divide by the value of the previous year:
Percentage change for year 2: (34,023.73 - 18,529.75) / 18,529.75 = 83.54%
Percentage change for year 3: (62,366.94 - 34,023.73) / 34,023.73 = 83.54%
Percentage change for year 4: (114,211.25 - 62,366.94) / 62,366.94 = 83.54%
For account B, we can see from the table that the percentage change after each year is different. To calculate the percentage change for each year, we divide the difference between the values by the value of the previous year and multiply by 100:
Percentage change for year 2: (18,794.10 - 9,638) / 9,638 * 100 = 94.89%
Percentage change for year 3: (36,648.50 - 18,794.10) / 18,794.10 * 100 = 94.89%
Percentage change for year 4: (71,464.58 - 36,648.50) / 36,648.50 * 100 = 94.89%
From the calculations, we can see that both account A and account B have the exact same percentage change in the amount of money over the previous year for all years. Therefore, there is no greater percentage change in either account. They both have a consistent percentage change of 83.54% or 94.89%.
Thank you so much!!! :)
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