Other Exponent Applications - Algebra 2

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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?

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Answer

Apply the compound interest formula:

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

We set final principal original principal , 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?

Tap to reveal answer

Answer

Set and solve for :

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

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

$\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?

Tap to reveal answer

Answer

We set final principal original principal , 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?

Tap to reveal answer

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.

Tap to reveal answer

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:

$y\cdot y= A$

← Didn't Know|Knew It →

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?

Tap to reveal answer

Answer

Apply the compound interest formula:

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

We set final principal original principal , 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.

← Didn't Know|Knew It →

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?

Tap to reveal answer

Answer

Set and solve for :

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

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

$\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.

← Didn't Know|Knew It →

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?

Tap to reveal answer

Answer

We set final principal original principal , 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.

← Didn't Know|Knew It →

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?

Tap to reveal answer

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.

← Didn't Know|Knew It →

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.

Tap to reveal answer

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:

$y\cdot y= A$

← Didn't Know|Knew It →

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?

Tap to reveal answer

Answer

We set final principal original principal , 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.

← Didn't Know|Knew It →

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?

Tap to reveal answer

Answer

Apply the compound interest formula:

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

We set final principal original principal , 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.

← Didn't Know|Knew It →

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?

Tap to reveal answer

Answer

Set and solve for :

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

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

$\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.

← Didn't Know|Knew It →

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?

Tap to reveal answer

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.

← Didn't Know|Knew It →

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.

Tap to reveal answer

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:

$y\cdot y= A$

← Didn't Know|Knew It →

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?

Tap to reveal answer

Answer

We set final principal original principal , 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.

← Didn't Know|Knew It →

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?

Tap to reveal answer

Answer

Apply the compound interest formula:

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

We set final principal original principal , 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.

← Didn't Know|Knew It →

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?

Tap to reveal answer

Answer

Set and solve for :

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

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

$\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.

← Didn't Know|Knew It →

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?

Tap to reveal answer

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.

← Didn't Know|Knew It →

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.

Tap to reveal answer

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:

$y\cdot y= A$

← Didn't Know|Knew It →

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