Question 1

What is the rate on an investment that doubles $2520 in 17 years? Assume interest is compounded quarterly.
A) 5.1%
B) 4.1%
C) 3.1%
D) 2.0%

Accepted Answer

The formula for compound interest is $A = P(1 + \frac{r}{n})^{nt}$, where $A$ is the amount of money accumulated after n years, including interest, $P$ is the principal amount (the initial amount of money), $r$ is the annual interest rate (decimal), $n$ is the number of times that interest is compounded per year, and $t$ is the time the money is invested for in years. In this case, we want to find $r$ when $A = 2P = 5040$, $P = 2520$, $n = 4$ (since interest is compounded quarterly), and $t = 17$. Rearranging the formula to solve for $r$ and substituting the given values, we find that the rate is approximately 4.1%.

Question 2

How long must $4400 be in a bank at 3% compounded annually to become $5913.23? (Round to the nearest year.)
A) 12 yr
B) 9 yr
C) 11 yr
D) 10 yr

Accepted Answer

The formula for compound interest is $A = P(1 + \frac{r}{n})^{nt}$, where $A$ is the amount of money accumulated after $n$ years, including interest, $P$ is the principal amount (the initial amount of money), $r$ is the annual interest rate (decimal), $n$ is the number of times that interest is compounded per year, and $t$ is the time the money is invested for in years. In this case, since the interest is compounded annually, $n = 1$, and we are given $A = 5913.23$, $P = 4400$, and $r = 0.03$ (3% as a decimal). We need to solve for $t$.Rearranging the formula to solve for $t$, we get:$$t = \frac{\log(A/P)}{\log(1 + r)}$$Substituting the given values:$$t = \frac{\log(5913.23/4400)}{\log(1 + 0.03)}$$$$t = \frac{\log(1.344375)}{\log(1.03)}$$$$t \approx 10$$Therefore, it takes approximately 10 years for $4400 to grow to $5913.23 at an annual compound interest rate of 3%.

Question 3

Find the number of years for $14,100 to grow to $16,600 at 3% compounded semiannually. Round to the nearest tenth of a year.
A) 5.4 yr
B) 6.6 yr
C) 4.6 yr
D) 5.5 yr

Accepted Answer

The formula for compound interest is $A = P(1 + \frac{r}{n})^{nt}$, where $A$ is the amount of money accumulated after $n$ years, including interest, $P$ is the principal amount (the initial amount of money), $r$ is the annual interest rate (decimal), $n$ is the number of times that interest is compounded per year, and $t$ is the time the money is invested for in years. Here, $A = \$16,600$, $P = \$14,100$, $r = 3\% = 0.03$, and $n = 2$ (since the interest is compounded semiannually). We need to solve for $t$.Rearranging the formula to solve for $t$, we get:$$t = \frac{\log(A/P)}{n \cdot \log(1 + r/n)}$$Substituting the given values:$$t = \frac{\log(16600/14100)}{2 \cdot \log(1 + 0.03/2)}$$$$t = \frac{\log(1.1773)}{2 \cdot \log(1.015)}$$$$t \approx \frac{0.0706}{2 \cdot 0.0064}$$$$t \approx \frac{0.0706}{0.0128}$$$$t \approx 5.5156$$Rounded to the nearest tenth of a year, $t \approx 5.5$ years.

Question 4

Solve the equation and express the solution in exact form.
-$$\log _ { 4 } \left( \log _ { 4 } x \right) = 1$$
A) {16}
B) {4}
C) {256}
D) {8}

Accepted Answer

The answer of Solve the equation and express the solution...

Question 5

What is the rate on an investment that triples $9465 in 35 years? Assume interest is compounded monthly.
A) 3.1%
B) 2.6%
C) 3.9%
D) 2.1%

Accepted Answer

The rate can be found using the formula for compound interest: A = P(1 + r/n)^(nt), where A is the amount of money accumulated after n years, including interest, P is the principal amount (the initial amount of money), r is the annual interest rate (decimal), n is the number of times that interest is compounded per year, and t is the time the money is invested for in years. Here, A = 3*9465, P = 9465, n = 12 (compounded monthly), and t = 35. Rearranging the formula to solve for r gives us the correct interest rate. Plugging the values into the formula and solving for r gives an interest rate of approximately 3.1%.

Question 6

How long will it take for $800 to grow to $1300 at an interest rate of 5.6% if the interest is compounded quarterly? Round the number of years to the nearest hundredth.
A) 8.91 yr
B) 34.92 yr
C) 8.73 yr
D) 29.22 yr

Accepted Answer

The formula for compound interest is $A = P(1 + \frac{r}{n})^{nt}$, where $A$ is the amount of money accumulated after n years, including interest, $P$ is the principal amount (the initial amount of money), $r$ is the annual interest rate (decimal), $n$ is the number of times that interest is compounded per year, and $t$ is the time the money is invested for in years. Here, $A = 1300$, $P = 800$, $r = 0.056$, and $n = 4$ (since the interest is compounded quarterly). We solve for $t$. Rearranging the formula to solve for $t$, we get $t = \frac{\log(A/P)}{n\log(1 + r/n)}$. Substituting the given values, $t = \frac{\log(1300/800)}{4\log(1 + 0.056/4)}$, which approximately equals 8.73 years.

Question 7

Solve the equation and express the solution in exact form.
-$$\log _ { 2 } x ^ { 2 } = \left( \log _ { 2 } x \right) ^ { 2 }$$
A) {0, 2}
B) {4}
C) {1, 2}
D) {1, 4}

Accepted Answer

Let $y = \log_2 x$. The equation becomes $y^2 = (\log_2 x)^2$, which simplifies to $y^2 = 2y$. Solving for $y$, we get $y^2 - 2y = 0$, leading to $y(y - 2) = 0$. Thus, $y = 0$ or $y = 2$. Converting back to $x$, when $y = 0$, $x = 2^0 = 1$, and when $y = 2$, $x = 2^2 = 4$. Therefore, the solution set is $\{1, 4\}$.

Question 8

Solve for the indicated variable.
-$A = \frac { t s } { 1 - ( 1 + z ) ^ { - B } }$, for B
A) $B = \frac { \ln A - \ln ( A - t s ) } { \ln ( 1 + z ) }$
B) $B = \ln \frac { t } { ( A - A s ) ( 1 + z ) }$
C) $B = \frac { \ln t } { \ln [ ( A - A s ) ( 1 - z ) ] }$
D) $B = \ln \frac { A } { ( A - t s ) ( 1 + z ) }$

Accepted Answer

To solve for $B$, we start with the given equation $A = \frac { t s } { 1 - ( 1 + z ) ^ { - B } }$ and rearrange it to isolate $B$. First, multiply both sides by the denominator and then divide by $A$ to get $1 - (1+z)^{-B} = \frac{ts}{A}$. Rearrange to get $(1+z)^{-B} = 1 - \frac{ts}{A}$, which simplifies to $(1+z)^{-B} = \frac{A-ts}{A}$. Taking the natural logarithm of both sides gives $-B \ln(1+z) = \ln\left(\frac{A-ts}{A}\right)$, which can be rearranged to $B = \frac{\ln A - \ln(A-ts)}{\ln(1+z)}$, matching choice A.

Question 9

Solve the equation and express the solution in exact form.
-$\ln \mathrm { e } ^ { \mathrm { x } } - \ln \mathrm { e } ^ { 7 } = \ln \mathrm { e } ^ { 4 }$
A) $\{ 3 \}$
B) $\{ - 3 \}$
C) $\{ 11 \}$
D) $\{ 28 \}$

Accepted Answer

The answer of Solve the equation and express the solution...

Question 10

Use your graphing calculator to find how long it will take for $9000 invested at 3.325% per year compounded daily to triple in value. Find the answer to the nearest year. Use 365 days for a year.
A) 44,219 yr
B) 4,402,084 yr
C) 33 yr
D) 0 yr

Accepted Answer

To find the time it takes for an investment to triple in value with daily compounding interest, we use the formula $A = P(1 + \frac{r}{n})^{nt}$, where $A$ is the amount of money accumulated after n years, including interest, $P$ is the principal amount (the initial amount of money), $r$ is the annual interest rate (decimal), $n$ is the number of times that interest is compounded per year, and $t$ is the time the money is invested for in years. We want $A = 3P$, $r = 0.03325$, $n = 365$, and $P = 9000$. Solving for $t$ gives us approximately 33 years.

Question 11

Solve for the indicated variable.
-$\mathrm { P } = 50,000 \cdot \mathrm { e } ^ { \mathrm { s } / 4 }$, for $\mathrm { s }$
A) $s = 4 \cdot \ln \frac { P } { 50,000 }$
B) $\mathrm { s } = \ln ( 50 \mathrm { P } )$
C) $\mathrm { s } = 50 \cdot \ln \mathrm { P }$
D) $s = 4 \cdot \ln \frac { \mathrm { P } } { 4000 }$

Accepted Answer

To solve for $s$, divide both sides by 50,000, then take the natural logarithm ($\ln$) of both sides, resulting in $s = 4 \cdot \ln \frac { P } { 50,000 }$.

Question 12

Solve for the indicated variable.
-$v = p - k \ln t$, for $t$
A) $t = \ln \frac { p - v } { k }$
B) $t = e ( p - v ) / k$
C) $\mathrm { t } = \frac { \mathrm { p } - \mathrm { v } } { \mathrm { k } }$
D) $t = e ^ { ( v - p ) / k }$

Accepted Answer

The answer of Solve for the indicated variable.
-\(v = p...

Question 13

At what interest rate must $5000 be compounded annually to equal $7116.56 after 9 yr? (Round to the nearest percent.)
A) 3%
B) 4%
C) 5%
D) 6%

Accepted Answer

The formula for compound interest is A = P(1 + r/n)^(nt), where A is the amount of money accumulated after n years, including interest, P is the principal amount, r is the annual interest rate (decimal), n is the number of times that interest is compounded per year, and t is the time the money is invested for in years. In this case, we are looking for r, the interest rate, when A = $7116.56, P = $5000, n = 1 (since the interest is compounded annually), and t = 9 years. Rearranging the formula to solve for r gives us r = [(A/P)^(1/nt) - 1]. Substituting the given values and solving for r gives us an interest rate that, when rounded to the nearest percent, is approximately 4%.

Question 14

Solve for the indicated variable.
-$\mathrm { I } - \mathrm { I } _ { 0 } = \left( \mathrm { I } _ { 1 } - \mathrm { I } _ { 0 } \right) 10 ^ { - \mathrm { kt } }$, for $\mathrm { t }$
A) $\mathrm { t } = - \frac { 1 } { \mathrm { k } } \log \left( \mathrm { I } - \mathrm { I } _ { 1 } \right)$
B) $t = - \frac { 1 } { k } \log \frac { \mathrm { I } } { \mathrm { I } }$
C) $t = - \frac { I - I _ { 0 } } { k \left( I _ { 1 } - I _ { 0 } \right) }$
D) $t = - \frac { 1 } { k } \log \frac { \mathrm { I } - \mathrm { I } _ { 0 } } { \mathrm { I } _ { 1 } - \mathrm { I } _ { 0 } }$

Accepted Answer

To solve for $t$, first divide both sides by $(I_1 - I_0)$ to isolate the exponential term, then apply the logarithm to both sides to solve for $t$, resulting in $t = - \frac { 1 } { k } \log \frac { I - I_0 } { I_1 - I_0 }$.

Question 15

Solve for the indicated variable.
-$\log _ { c } ( z + s ) = \log _ { c } ( 2 z - 4 )$, for $s$
A) $s = \frac { z } { 4 }$
B) $s = \frac { \log _ { c } z } { 4 }$
C) $s = z - 4$
D) $s = \log _ { c } ( z - 4 )$

Accepted Answer

Given $\log_c(z + s) = \log_c(2z - 4)$, since the bases of the logarithms are the same, the arguments must be equal. Therefore, $z + s = 2z - 4$. Solving for $s$, we subtract $z$ from both sides to get $s = z - 4$.

Question 16

Solve the equation and express the solution in exact form.
-$\log _ { 4 } ( x + 1 ) + \log _ { 4 } ( x - 1 ) = 2$
A) $\left\{ \frac { 3 } { 2 } \right\}$
B) $\{ 17 \}$
C) $\{ \sqrt { 17 } \}$
D) $\{ 8 \}$

Accepted Answer

Using the property of logarithms that allows combining two logs with the same base into a single log (logarithm of a product), the equation becomes $\log_4((x+1)(x-1)) = 2$. Simplifying inside the log gives $\log_4(x^2-1) = 2$. To solve for $x$, convert the logarithmic equation to its exponential form: $4^2 = x^2 - 1$, which simplifies to $16 = x^2 - 1$. Solving for $x^2$ gives $x^2 = 17$, and thus $x = \sqrt{17}$.

Question 17

Solve the equation and express the solution in exact form.
-$\ln x - \ln ( x - 3 ) = \ln 7$
A) $\varnothing$
B) $\left\{ \frac { 7 } { 2 } \right\}$
C) $\{ 4 \}$
D) $\left\{ \frac { 3 \ln 7 } { \ln 7 - 1 } \right\}$

Accepted Answer

The answer of Solve the equation and express the solution...

Question 18

Solve the equation and express the solution in exact form.
-$$\log _ { 2 } \sqrt { 2 x ^ { 2 } } = \frac { 11 } { 2 }$$
A) $\{ - 16,16 \}$
B) $\{ 32 \}$
C) $\{ - 32,32 \}$
D) $\{ - 1024,1024 \}$

Accepted Answer

The equation simplifies to $ \log_2 |x| = \frac{11}{2} $, which means $ |x| = 2^{\frac{11}{2}} = 32 $. Thus, $ x = \pm 32 $.

Question 19

Solve for the indicated variable.
-$\mathrm { T } = \mathrm { T } _ { 0 } \mathrm { e } ^ { \mathrm { kt } / 300 }$, for $\mathrm { t }$
A) $t = \frac { k } { 300 } \ln \frac { T } { T _ { 0 } }$
B) $t = \frac { 300 } { k } \ln \left( T - T _ { 0 } \right)$
C) $t = \frac { 300 \mathrm {~T} } { \mathrm { kT } _ { 0 } }$
D) $t = \frac { 300 } { k } \ln \frac { T } { T _ { 0 } }$

Accepted Answer

To solve for $t$, divide both sides by $T_0$, then take the natural logarithm of both sides to get rid of the exponential, resulting in $t = \frac{300}{k} \ln \frac{T}{T_0}$.

Question 20

Solve for the indicated variable.
-$I = \frac { \mathrm { w } } { \mathrm { P } } \left( 1 - \mathrm { e } ^ { - \mathrm { as } / 3 } \right)$, for $\mathrm { s }$
A) $s = - \frac { 3 \ln \left( 1 - I P _ { W } \right) } { a }$
B) $s = - \frac { 3 } { a } \ln \left( 1 - \frac { \mathrm { IP } } { \mathrm { w } } \right)$
C) $\mathrm { s } = \frac { \mathrm { IwP } \cdot \ln ( 1 - \mathrm { a } ) } { 3 }$
D) $\mathrm { s } = \frac { 3 \cdot \ln ( 1 - \mathrm { a } ) } { \mathrm { IwP } }$

Accepted Answer

To solve for $s$, we first isolate the exponential term: $I = \frac { w } { P } \left( 1 - \mathrm { e } ^ { - \mathrm { as } / 3 } \right)$ implies $1 - \frac{IP}{w} = e^{-as/3}$. Taking the natural logarithm of both sides gives $\ln\left(1 - \frac{IP}{w}\right) = -\frac{as}{3}$, which rearranges to $s = -\frac{3}{a} \ln\left(1 - \frac{IP}{w}\right)$.

Quiz 5: Inverse, Exponential, and Logarithmic Functions

What is the rate on an investment that doubles $2520 in 17 years? Assume interest is compounded quarterly.

How long must $4400 be in a bank at 3% compounded annually to become $5913.23? (Round to the nearest year.)

Find the number of years for $14,100 to grow to $16,600 at 3% compounded semiannually. Round to the nearest tenth of a year.

Solve the equation and express the solution in exact form. - $\log _ { 4 } \left( \log _ { 4 } x \right) = 1$

What is the rate on an investment that triples $9465 in 35 years? Assume interest is compounded monthly.

How long will it take for $800 to grow to $1300 at an interest rate of 5.6% if the interest is compounded quarterly? Round the number of years to the nearest hundredth.

Solve the equation and express the solution in exact form. - $\log _ { 2 } x ^ { 2 } = \left( \log _ { 2 } x \right) ^ { 2 }$

Solve for the indicated variable. - $A = \frac { t s } { 1 - ( 1 + z ) ^ { - B } }$ , for B

Solve the equation and express the solution in exact form. - $\ln \mathrm { e } ^ { \mathrm { x } } - \ln \mathrm { e } ^ { 7 } = \ln \mathrm { e } ^ { 4 }$

Use your graphing calculator to find how long it will take for $9000 invested at 3.325% per year compounded daily to triple in value. Find the answer to the nearest year. Use 365 days for a year.

Solve for the indicated variable. - $\mathrm { P } = 50,000 \cdot \mathrm { e } ^ { \mathrm { s } / 4 }$ , for $\mathrm { s }$

Solve for the indicated variable. - $v = p - k \ln t$ , for $t$

At what interest rate must $5000 be compounded annually to equal $7116.56 after 9 yr? (Round to the nearest percent.)

Solve for the indicated variable. - $\mathrm { I } - \mathrm { I } _ { 0 } = \left( \mathrm { I } _ { 1 } - \mathrm { I } _ { 0 } \right) 10 ^ { - \mathrm { kt } }$ , for $\mathrm { t }$

Solve for the indicated variable. - $\log _ { c } ( z + s ) = \log _ { c } ( 2 z - 4 )$ , for $s$

Solve the equation and express the solution in exact form. - $\log _ { 4 } ( x + 1 ) + \log _ { 4 } ( x - 1 ) = 2$

Solve the equation and express the solution in exact form. - $\ln x - \ln ( x - 3 ) = \ln 7$

Solve the equation and express the solution in exact form. - $\log _ { 2 } \sqrt { 2 x ^ { 2 } } = \frac { 11 } { 2 }$

Solve for the indicated variable. - $\mathrm { T } = \mathrm { T } _ { 0 } \mathrm { e } ^ { \mathrm { kt } / 300 }$ , for $\mathrm { t }$

Solve for the indicated variable. - $I = \frac { \mathrm { w } } { \mathrm { P } } \left( 1 - \mathrm { e } ^ { - \mathrm { as } / 3 } \right)$ , for $\mathrm { s }$

Quiz 5: Inverse, Exponential, and Logarithmic Functions

What is the rate on an investment that doubles $2520 in 17 years? Assume interest is compounded quarterly.

How long must $4400 be in a bank at 3% compounded annually to become $5913.23? (Round to the nearest year.)

Find the number of years for $14,100 to grow to $16,600 at 3% compounded semiannually. Round to the nearest tenth of a year.

Solve the equation and express the solution in exact form. - log⁡4(log⁡4x) =1\log _ { 4 } \left( \log _ { 4 } x \right) = 1log4​(log4​x) =1

What is the rate on an investment that triples $9465 in 35 years? Assume interest is compounded monthly.

How long will it take for $800 to grow to $1300 at an interest rate of 5.6% if the interest is compounded quarterly? Round the number of years to the nearest hundredth.

Solve the equation and express the solution in exact form. - log⁡2x2=(log⁡2x) 2\log _ { 2 } x ^ { 2 } = \left( \log _ { 2 } x \right) ^ { 2 }log2​x2=(log2​x) 2

Solve for the indicated variable. - A=ts1−(1+z) −BA = \frac { t s } { 1 - ( 1 + z ) ^ { - B } }A=1−(1+z) −Bts​ , for B

Solve the equation and express the solution in exact form. - ln⁡ex−ln⁡e7=ln⁡e4\ln \mathrm { e } ^ { \mathrm { x } } - \ln \mathrm { e } ^ { 7 } = \ln \mathrm { e } ^ { 4 }lnex−lne7=lne4

Use your graphing calculator to find how long it will take for $9000 invested at 3.325% per year compounded daily to triple in value. Find the answer to the nearest year. Use 365 days for a year.

Solve for the indicated variable. - P=50,000⋅es/4\mathrm { P } = 50,000 \cdot \mathrm { e } ^ { \mathrm { s } / 4 }P=50,000⋅es/4 , for s\mathrm { s }s

Solve for the indicated variable. - v=p−kln⁡tv = p - k \ln tv=p−klnt , for ttt

At what interest rate must $5000 be compounded annually to equal $7116.56 after 9 yr? (Round to the nearest percent.)

Solve for the indicated variable. - I−I0=(I1−I0) 10−kt\mathrm { I } - \mathrm { I } _ { 0 } = \left( \mathrm { I } _ { 1 } - \mathrm { I } _ { 0 } \right) 10 ^ { - \mathrm { kt } }I−I0​=(I1​−I0​) 10−kt , for t\mathrm { t }t

Solve for the indicated variable. - log⁡c(z+s) =log⁡c(2z−4) \log _ { c } ( z + s ) = \log _ { c } ( 2 z - 4 ) logc​(z+s) =logc​(2z−4) , for sss

Solve the equation and express the solution in exact form. - log⁡4(x+1) +log⁡4(x−1) =2\log _ { 4 } ( x + 1 ) + \log _ { 4 } ( x - 1 ) = 2log4​(x+1) +log4​(x−1) =2

Solve the equation and express the solution in exact form. - ln⁡x−ln⁡(x−3) =ln⁡7\ln x - \ln ( x - 3 ) = \ln 7lnx−ln(x−3) =ln7

Solve the equation and express the solution in exact form. - log⁡22x2=112\log _ { 2 } \sqrt { 2 x ^ { 2 } } = \frac { 11 } { 2 }log2​2x2​=211​

Solve for the indicated variable. - T=T0ekt/300\mathrm { T } = \mathrm { T } _ { 0 } \mathrm { e } ^ { \mathrm { kt } / 300 }T=T0​ekt/300 , for t\mathrm { t }t

Solve for the indicated variable. - I=wP(1−e−as/3) I = \frac { \mathrm { w } } { \mathrm { P } } \left( 1 - \mathrm { e } ^ { - \mathrm { as } / 3 } \right) I=Pw​(1−e−as/3) , for s\mathrm { s }s

Solve the equation and express the solution in exact form. - $\log _ { 4 } \left( \log _ { 4 } x \right) = 1$

Solve the equation and express the solution in exact form. - $\log _ { 2 } x ^ { 2 } = \left( \log _ { 2 } x \right) ^ { 2 }$

Solve for the indicated variable. - $A = \frac { t s } { 1 - ( 1 + z ) ^ { - B } }$ , for B

Solve the equation and express the solution in exact form. - $\ln \mathrm { e } ^ { \mathrm { x } } - \ln \mathrm { e } ^ { 7 } = \ln \mathrm { e } ^ { 4 }$

Solve for the indicated variable. - $\mathrm { P } = 50,000 \cdot \mathrm { e } ^ { \mathrm { s } / 4 }$ , for $\mathrm { s }$

Solve for the indicated variable. - $v = p - k \ln t$ , for $t$

Solve for the indicated variable. - $\mathrm { I } - \mathrm { I } _ { 0 } = \left( \mathrm { I } _ { 1 } - \mathrm { I } _ { 0 } \right) 10 ^ { - \mathrm { kt } }$ , for $\mathrm { t }$

Solve for the indicated variable. - $\log _ { c } ( z + s ) = \log _ { c } ( 2 z - 4 )$ , for $s$

Solve the equation and express the solution in exact form. - $\log _ { 4 } ( x + 1 ) + \log _ { 4 } ( x - 1 ) = 2$

Solve the equation and express the solution in exact form. - $\ln x - \ln ( x - 3 ) = \ln 7$

Solve the equation and express the solution in exact form. - $\log _ { 2 } \sqrt { 2 x ^ { 2 } } = \frac { 11 } { 2 }$

Solve for the indicated variable. - $\mathrm { T } = \mathrm { T } _ { 0 } \mathrm { e } ^ { \mathrm { kt } / 300 }$ , for $\mathrm { t }$

Solve for the indicated variable. - $I = \frac { \mathrm { w } } { \mathrm { P } } \left( 1 - \mathrm { e } ^ { - \mathrm { as } / 3 } \right)$ , for $\mathrm { s }$