Applying Exponents - Algebra II

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Question

Peter opens a savings account on his $21^{st}$ t birthday. He makes a deposit of . The account earns percent interest, compounded annually. Peter plans to take the money out when he is years old. If he doesn't make any deposits or withdrawals until then, how much money will be in the account?

Answer

The formula for calculating compount interest is as follows:

$FV = PV\cdot (1+r)^n$

where

= future value

= present value

= interest rate

= number of times the interest is compounded

In this problem, the present value of the money is $5000, and the interest rate is 7%. If Peter takes the money out when he is 50, it would have been compounded 29 times (once per year). Therefore:

$FV = $5000\cdot (1+.07)^{29}$

${\color{Red} FV = $35,571.29}$

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Question

Catherine invests $3500 in an investment account. The account earns 10% interest, compounded quarterly. After 5 years, how much money will she have?

Answer

The formula for calculating the future value of an interest earning account is

$FV = PV(1+\frac{r}{n})^{tn}$ ,

where

= future value,

= present value,

= annual interest rate,

= number of times the interest is compounded per year, and

= the number of years that have passed.

The problem asks for the amount of money in the account after 5 years, with 10% interested compounded four times per year (quarterly).

Plug in the given quantities and simplify:

$FV=$3500 (1+.025)^{5\cdot 4}$

$FV=3500(1.025)^{20}$

${\color{Red} FV=$5,735.15}$

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Question

Felicia put money in a saving account with a 5% interest rate, compounded annually. After five years, she had $10,000. How much was her initial investment?

Answer

The formula for finding the future value of an investment is

,

where

= future value,

= present value,

= interest rate, and

= number of times interest is compounded.

Plug in the given numbers and solve for the present value:

$PV=\frac{10,000}{1.05^5}$

${\color{Red} PV=7,835.26}$

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Question

Round the answer to two decimals.

Anthony put , in his savings account today. The bank pays interest of every year.

How much does he have in his savings account after years?

Answer

The formula for computing interest is:

Beginning Amount x ((1 + rate)^number of years) = Ending Amount After number of years

Make sure to convert the rate from percent to number: 3% = 0.03

$E=6000 (1+0.03)^{15}$

So the answer is

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Question

Jamie deposits $5000 into an account at ABC bank. The account will earn a 4% interest rate compounded yearly. Jamie would like to withdraw the accumulated amount after 5 years and close the account. How much money would Jamie withdraw after 5 years? (Round your answer to the nearest dollar)

Answer

Initial amount = 5000

The account earns 4% compounded yearly ===> Each $1.00 will grow into $1.04.

Growth rate = 1.04

Jamie will withdraw the money after 5 years. Since the interest is compounded yearly, the number of periods is equal to the number of years the money will be in the account.

number of periods = 5

From the above information, we can calculate the amount accumulated (or final amount) after 5 years using the following formula:

final amount = initial amount * (growth rate)number of periods

$5000\times(1.04)^{5}= 6083.2645\approx 6083$

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Question

For coninuous compound interest:

$P = Ce^{rt}$

Where

$P = \text{ final principal}$

$C = \text{ initial deposit}$

$r = \text{ interest rate}$

$t =\text{ number of years of investment}$

If an initial deposit of is continuously compounded at a rate of for years, what will be the final principal value to the nearest dollar?

Answer

Using the equation for continuous compound interest and the given information, we get

$P = Ce^{rt}$

$P = \text{ final principal}$

$r = \text{ 0.06}$

$P = 250e^{\left ( .06\cdot 3\right )} =250e^{.18} = 250 \cdot 1.1972 = 299$

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Question

Remember $A = P(1+\frac{r}{n})^{nt}$

If an account has a starting principle P = $5,000, an interest rate r = 12% or 0.12, compounded annually, how much money should there be after five years? Assume no money has been added or taken out of the account since it was opened.

Answer

$A = P(1+\frac{r}{n})^{nt}$ is the compound interest formula where

P = Initial deposit = 5000

r = Interest rate = 0.12

n = Number of times interest is compounded per year = 1

t = Number of years that have passed = 5

Round to the nearest cent or hundredth is .

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Question

Julio invests $5000 into an account with a 2.5% interest rate, compounded quarterly. What is his account balance after 1 year (rounded to the nearest cent)?

Answer

To determine Julio's account balance, we must use the interest formula given below:

$P(1+\frac{r}{n})^{nt}$

where P is his principal (initial) investment, r is the interest rate (as a decimal), n is the number of times the interest is compounded, and t is the amount of time elapsed.

Plugging in all of our given information into the above formula - knowing that quarterly means four times a year - we get

$5000(1+\frac{0.025}{4})^{4\cdot 1}=5126.1767...\approx5126.18$

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Question

Martisha invests $2000 into an account with continuously compounded interest. The account has an interest rate of 2.5%. Find the balance of the account after 2 years, rounded to the nearest cent.

Answer

To find the balance, B, of a continuously compounded interest account after a certain amount of time, we must use the following formula:

$B=Pe^{rt}$ , where P is the initial investment, r is the interest rate (as a decimal), and t is the amount of time being considered.

Plugging in all of the given information, we get

$B=2000e^{0.025\cdot 2}=2102.542193$

which rounded becomes $2102.54

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Question

How long will it take for Nikki to triple her initial investment into a continuously compounded interest account with an interest rate of 1.9%?

Answer

The formula to find the balance, B, of a continuously compounded interest account with interest rate, r, after a certain time, t, is given by

$B=Pe^{rt}$

To solve this problem, we need to know only the initial investment (P), our final balance (three times P) and the interest rate (expressed as a decimal), 0.019.

Plugging in our known information into the formula for continuously compounded interest, we get

$3P=Pe^{0.019t}$

We now solve for t:

$3=e^{0.019t}$

$\ln(3)=\ln(e^{0.019t})$

Exponentiating both sides allows us to get rid of the exponential:

$\ln3=0.019t$

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Question

Sheila wants to double her initial investment into a compounded interest account, with an interest rate of 4%. How long will this take, if the interest is compounded annually?

Answer

To determine the amount of time needed to double the initial investment - P - into a compound interest account, we simply plug in our given information into the formula:

$B=P(1+\frac{r}{n})^{nt}$

where B is the balance, P is the initial investment, r is the interest rate (as a decimal), n is the number of times the interest is compounded, and t is the time elapsed.

Now, because we are doubling P, our balance B becomes two times P:

$2P=P(1+\frac{0.04}{1})^{1\cdot t}$

Now, we can solve for P:

To bring the time variable down from being an exponent, we take the logarithm of both sides (common or natural):

$\ln2=t\cdot \ln1.04$

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Question

If a person deposits 300 dollars to a savings account, which earns one percent interest that is compounded annually, what is the balance after 60 years?

Answer

Write the formula for compound interest.

$A=P(1+\frac{r}{n})^{nt}$

$A= \textup{balance}$

$P=\textup{starting investment or principal} = $300$

$r=\textup{ rate} = 1% = 0.01$

$n= \textup{number of times compounded per year} =1$

$t= \textup{time in years} = 60$

Substitute all the known values into the formula.

$A=300(1+\frac{0.01}{1})^{(1)(60)} = 300(1.01)^{60} = $545$

The answer is:

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Question

Suppose Billy's has , invests the money at a bank at , compounded monthly. About how much will Billy have after 36 months?

Answer

Write the compound interest formula.

$A=P(1+\frac{r}{n})^{nt}$

where is the total, is the principal, is the rate, is the number of times compounded annually, and is the time in years.

$t=\frac{36}{12}=3$

Substitute all into the equation.

$A=300(1+\frac{0.01}{12})^{12\times 3}$

The answer is:

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Question

Ann deposits $30,000 into a savings account that pays 4.5% annual interest compounded quarterly. Assuming she neither deposits nor withdraws money, what is the amount of time it will take for her to have at least $50,000 in the account?

Answer

Apply the compound interest formula:

$A = A_{0} \left ( 1 + \frac{r}{n}\right ) ^{nt}$ .

We set final principal original principal $A_{0} = 30,000$ , interest rate , number of periods per year (quarterly). We solve for in the equation

$50,000=30,000 \left ( 1 + \frac{0.045}{4}\right ) ^{4t}$

$50,000=30,000 \left ( 1 .01125 \right ) ^{4t}$

$\frac{50,000}{30,000}=\frac{30,000 \left ( 1 .01125 \right ) ^{4t}}{30,000}$

$1.6667= \left ( 1 .01125 \right ) ^{4t}$

$\log 1.6667= \log \left ( 1 .01125 \right ) ^{4t}$

$\log 1.6667= 4t \log 1 .01125$

$t = \frac{\log 1.6667}{4\log 1 .01125} \approx \frac{0.2219}{4 \cdot 0.004859} \approx 11.4$

This is rounded up to the next quarter of a year, so the correct response is 11.5 years, or 11 years 6 months.

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Question

A biologist figures that the population of cane toads in a certain lake he is studying can be modeled by the equation

$P = 253 \cdot 1.04 ^{t}$ ,

where is the number of days elapsed in 2015. For example, represents January 1, represents January 2, and so forth.

If this model continues, in what month will the population of cane toads in the lagoon reach 5,000?

Answer

Set and solve for :

$5,000 = 253 \cdot 1.04^{t}$

$1.04 ^{t} = \frac{5,000 }{253 }$

$1.04^{t} \approx 19.7628$

$\ln 1.04^{t} \approx \ln 19.7628$

$t \ln 1.04 \approx \ln 19.7628$

$t \approx \frac{\ln 19.7628}{\ln 1.04 } \approx \frac{2.9838}{0.0392} \approx 76.1$

January and February have 59 days total; add March, and this is 90 days. The 76th day is in March.

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Question

Lucia deposits $40,000 into a savings account that pays 5.5% annual interest compounded continuously. Assuming she neither deposits nor withdraws money, how long will it take for her to have $60,000 in the account?

Answer

We set final principal original principal $A_{0} = 40,000$ , and interest rate . We solve for in the continuous compound interest formula:

$A = A_{0} e ^{rt}$

$60,000 =40,000 e ^{0.055t}$

$\frac{60,000 }{40,000 }=\frac{40,000 e ^{0.055t}}{40,000 }$

$1.5 = e ^{0.055t}$

$0.055t = \ln 1.5$

$t = \frac{\ln 1.5}{0.055} \approx \frac{0.4055}{0.055} \approx 7.37$

The correct response is therefore between 7 and 8 years.

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Question

A biologist figures that the population of cane toads in a certain lake he is studying can be modeled by the equation

$P = 318\cdot 1.07 ^{t}$ ,

where is the number of days elapsed in 2015. For example, represents January 1, represents January 2, and so forth.

Assuming that this has been the model for their growth throughout the previous year as well, in what month did the population hit 100 cane toads?

Answer

Set and solve for :

$100 = 318\cdot 1.07 ^{t}$

$\frac{100}{318} = 1.07 ^{t}$

$1.07 ^{t} \approx 0.3145$

$\ln 1.07 ^{t} \approx \ln 0.3145$

$t \ln 1.07 \approx \ln 0.3145$

$t \approx \frac{\ln 0.3145}{ \ln 1.07} \approx \frac{-1.1568}{0.0677} \approx -17.1$

17 days before January 1 was in December of 2014.

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Question

A company is constructing a wall with 4-sides, all sides are of equal length.

Write an equation using exponents to calculate the area of the wall. Use as the length and height.

Answer

The formula to find area is:

$= length\cdot width=area$ is correct.

In our case our length is equal to our width which is .

Substituting our values into our equation we get:

$y\cdot y= A$

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Question

Sceintists recently discovered a new type of metal compound. They have roughly 15 grams of this compound, which has a half life of 16 hours. Approximately how much of this substance will the scientists have in 24 hours?

Answer

Recall the radioactive decay formula:

$Q_{(t)}=Q_{(o)} (^{-r\cdot t})$

The half life formula is:

$r=\frac{ln2}{(h)}$ , where is the half life.

Plug in the given half life:

$r=\frac{ln2}{(16)}\approx 0.0433216$

Plug this value into the radioactive decay formula:

$Q_{(24)}=15 (^{-0.0433216\cdot 24})\approx 0.06$ grams

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Question

Over the past few years, the number of students enrolled at a certain university has been decreasing. Each year there is a 12% decrease in student enrollement. Currently, 14,286 students are enrolled. If this trend continues, how many students will be enrolled in 5 years?

Answer

This is an exponential decay problem. The formula for exponential decay is:

Where

= future value

= present value

= rate of decay

= number of periods

This problem requests the number of students five years in the future. The rate of decay is twelve percent. Therefore:

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