#1 2000* 1.04^3
#3. 2000 * 1.04^x = 2500
1.04^x = 1.25
x = ln1.25/ln1.04
1.) How much money will be in the account after 3 years?
2.) How much money will be in the account after 18 years?
3.) How many years will it take for the account to contain $2,500?
4.) How many years will it take for the account to contain $3,000?
#3. 2000 * 1.04^x = 2500
1.04^x = 1.25
x = ln1.25/ln1.04
amount = 2000(1.04)^3 = ....
#2
same thing
#3.
the unknown is the time in our formula
2000(1.04)^t = 2500
1.04^t = 1.25
take log of both sides
log[ (1.04^t] = log 1.25
use rules of logs
t log 1.04 = log 1.25
t = log 1.25/log 1.04 = appr 5.7 years
#4
same thing
A = P(1 + r/n)^(nt)
where:
A = the future value of the investment/amount in the account
P = the principal amount (initial deposit)
r = annual interest rate (as a decimal)
n = number of times interest is compounded per year
t = number of years
In this case:
P = $2,000
r = 4% = 0.04 (as a decimal)
n = 1 (interest is compounded annually)
1.) To find how much money will be in the account after 3 years:
t = 3
A = 2000(1 + 0.04/1)^(1*3)
A = 2000(1 + 0.04)^3
A = 2000(1.04)^3
A = 2000(1.124864)
A ≈ $2,249.73
Therefore, there will be approximately $2,249.73 in the account after 3 years.
2.) To find how much money will be in the account after 18 years:
t = 18
A = 2000(1 + 0.04/1)^(1*18)
A = 2000(1 + 0.04)^18
A = 2000(1.04)^18
A ≈ $3,207.13
Therefore, there will be approximately $3,207.13 in the account after 18 years.
3.) To find how many years it will take for the account to contain $2,500:
A = $2,500
2,500 = 2000(1 + 0.04/1)^(1*t)
2,500 = 2000(1.04)^t
1.25 = 1.04^t
To solve for t, we can take the logarithm of both sides. Let's use the natural logarithm (ln) in this case:
ln(1.25) = ln(1.04^t)
ln(1.25) = t * ln(1.04)
Now, we can solve for t:
t = ln(1.25) / ln(1.04)
t ≈ 9.11
Therefore, it will take approximately 9.11 years for the account to contain $2,500.
4.) To find how many years it will take for the account to contain $3,000:
A = $3,000
3,000 = 2000(1 + 0.04/1)^(1*t)
3,000 = 2000(1.04)^t
Again, we will take the logarithm of both sides:
ln(3,000) = ln(2000(1.04)^t)
ln(3,000) = t * ln(1.04)
Now, solve for t:
t = ln(3,000) / ln(1.04)
t ≈ 11.42
Therefore, it will take approximately 11.42 years for the account to contain $3,000.
A = P(1 + r/n)^(nt)
Where:
A = the amount of money in the account after time t
P = the principal amount (initial deposit)
r = annual interest rate (expressed as a decimal)
n = number of times the interest is compounded per year
t = time in years
We are given the following information:
P = $2,000
r = 4% = 0.04 (decimal form)
1.) To find the amount of money in the account after 3 years:
A = 2000(1 + 0.04/1)^(1*3)
We use n = 1 because the interest is compounded annually.
A = 2000(1 + 0.04)^3
A = 2000(1.04)^3
A ≈ $2,249.82
Therefore, there will be approximately $2,249.82 in the account after 3 years.
2.) To find the amount of money in the account after 18 years:
A = 2000(1 + 0.04/1)^(1*18)
A = 2000(1 + 0.04)^18
A ≈ $3,017.40
Therefore, there will be approximately $3,017.40 in the account after 18 years.
3.) To find the number of years it will take for the account to contain $2,500:
We need to solve the equation A = 2000(1 + 0.04/1)^(1*t) for t. Rearranging the equation, we get:
2500 = 2000(1.04)^t
To solve for t, we can use logarithms. Take the natural logarithm (ln) of both sides of the equation:
ln(2500) = ln(2000(1.04)^t)
Using properties of logarithms, we can simplify the equation:
ln(2500) = ln(2000) + t*ln(1.04)
Now, divide both sides of the equation by ln(1.04):
t = (ln(2500) - ln(2000))/ln(1.04)
Calculating this equation will give us the number of years it takes for the account to contain $2,500.
4.) Likewise, to find the number of years it will take for the account to contain $3,000:
We need to solve the equation A = 2000(1 + 0.04/1)^(1*t) for t. Rearranging the equation, we get:
3000 = 2000(1.04)^t
Using logarithms, we can simplify the equation:
ln(3000) = ln(2000(1.04)^t)
Again, divide both sides of the equation by ln(1.04):
t = (ln(3000) - ln(2000))/ln(1.04)
Calculating this equation will give us the number of years it takes for the account to contain $3,000.