Me on joe deposits $1,500 in an account that pays 3% annual interest compounded continuously. a. how much will joe have in his account after 5 years? b. how long will it take joe to double his money? use natural logarithms and explain your answer.

ANSWERS:

Question A. $1742.751361739

Question B. 23.1 years

WORKINGS:

Interest compounded continuously is calculated with the formula:

A = Pe^(rt)

Where,

P = Principal amount

r = Interest rate

t = Time in years

Question A. How much will Joe have in his account after 5 years

Given that Joe deposits $1,500 in the account

P = 1500

Given that the account pays 3% annual interest

r = 3/100

r = 0.03

Since we want to know how much Joe will have after 5 years

t = 5

A = Pe^(rt)

A = 1500e^(0.03 * 5)

A = 1500e^(0.15)

Using definition of logarithm:
0.15 = ln(A / 1500)
0.15 = ln(A) - ln(1500)

Add ln(1500) to both sides of the equation

0.15 + ln(1500) = ln(A) - ln(1500) +  ln(1500)
0.15 + ln(1500) = ln(A)
ln(A) = 0.15 + ln(1500)

In (A) = 0.15 + 7.313220387

In (A) = 7.463220387

Solving for A

In (A) = 7.463220387

Therefore, A = e^(7.463220387)  ;     [e = 2.718281828]

A = 1742.751361739

Therefore, Joe have $1742.751361739 in his account after 5 years

Question B. How long will it take Joe to double his money?

A = Pe^(rt)

Given that Joe deposits $1,500 in the account

P = $1500

Given that the account pays 3% annual interest

r = 3/100

r = 0.03

Since we want to know how long it will take Joe to double his money

A = $1500 * 2 = $3000

Solving for t

A = Pe^(rt)

3000 = 1500e^(0.03t)

Using definition of logarithm

0.03t = In (3000/1500)
0.03t = ln(2)

Divide both sides of the equation by 0.03

0.03t/0.03 = ln(2)/0.03
t = ln(2) / 0.03

t = 0.693147181 / 0.03

t = 23.104906019

Therefore, it will take Joe 23.1 years to double his money.

answer: domain and range mean all possible values of x and y that could be substituted to the inequality. a solution means all possible values that make the inequality statement true. so while the range is still all real numbers, the solution is the shaded area, for which y will always be greater than the line of the function.

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