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Algebra II : Other Exponent Applications

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Example Questions

Example Question #1 : Other Exponent Applications

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, t= 1 represents January 1, t = 2 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?

Possible Answers:

May

March

June

February

April

Correct answer:

March

Explanation:

Set P = 5,000 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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Example Question #2 : Other Exponent Applications

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, t= 1 represents January 1, t = 2 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?

Possible Answers:

August 2014

September 2014

October 2014

December 2014

November 2014

Correct answer:

December 2014

Explanation:

Set P = 100 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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Example Question #3 : Other Exponent Applications

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?

Possible Answers:

Between 6 and 7 years

Between 9 and 10 years

Between 7 and 8 years

Between 8 and 9 years

Between 5 and 6 years

Correct answer:

Between 7 and 8 years

Explanation:

We set final principal A = 60,000 original principal $A_{0} = 40,000$ , and interest rate r = 0.055 . 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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Example Question #2 : Other Exponent Applications

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?

Possible Answers:

$12 \textup{ yrs } 6\textup{ mo}$

$11 \textup{ yrs }$

$11 \textup{ yrs } 6\textup{ mo}$

$13\textup{ yrs }$

$12\textup{ yrs }$

Correct answer:

$11 \textup{ yrs } 6\textup{ mo}$

Explanation:

Apply the compound interest formula:

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

We set final principal A = 50,000 original principal $A_{0} = 30,000$ , interest rate r = 0.045 , number of periods per year n = 4 (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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Example Question #801 : Mathematical Relationships And Basic Graphs

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.

Possible Answers:

$A=y\cdot y \cdot y \cdot y$

A=y^4

A=y+y

A=y^2

$A=y\cdot y$

Correct answer:

A=y^2

Explanation:

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:

$y\cdot y= A$

y^2=A

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Algebra II : Other Exponent Applications

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #1 : Other Exponent Applications

Example Question #2 : Other Exponent Applications

Example Question #3 : Other Exponent Applications

Example Question #2 : Other Exponent Applications

Example Question #801 : Mathematical Relationships And Basic Graphs

All Algebra II Resources

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