Question 1

Solve the problem.
-A fossilized leaf contains 12% of its normal amount of carbon 14. How old is the fossil (to the nearest year)? Use 5600 years as the half-life of carbon 14.

Accepted Answer

The age of the fossil can be calculated using the formula for radioactive decay, $N = N_0 \times (\frac{1}{2})^{\frac{t}{T}}$, where $N$ is the remaining amount of substance, $N_0$ is the initial amount of substance, $t$ is the time that has passed, and $T$ is the half-life of the substance. Given that $N/N_0 = 0.12$ (or 12%) and $T = 5600$ years, we can solve for $t$. Rearranging the formula gives us $t = \frac{T \times \log(N/N_0)}{\log(1/2)}$. Plugging in the values gives us $t \approx 17,099$ years.

Question 2

Solve the problem.
-How long will it take for an investment to double in value if it earns 4.25% compounded continuously? Round your answer to three decimal places.

Accepted Answer

The formula to calculate the time $ t $ for an investment to double when interest is compounded continuously is given by $ t = \frac{\ln(2)}{r} $, where $ r $ is the annual interest rate expressed as a decimal. For an interest rate of 4.25%, $ r = 0.0425 $. Plugging the values into the formula gives $ t = \frac{\ln(2)}{0.0425} \approx 16.309 $ years.

Question 3

Solve the problem.
-If Emery has $1100 to invest at 6% per year compounded monthly, how long will it be before he has $2100? If the compounding is continuous, how long will it be? (Round your answers to three decimal places.)

Accepted Answer

For monthly compounding, the formula 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 ($1100), r is the annual interest rate (6%, or 0.06), n is the number of times that interest is compounded per year (12 for monthly), and t is the time the money is invested for in years. For continuous compounding, the formula is A = Pe^(rt), where e is the base of the natural logarithm (approximately 2.71828). Solving each equation for t gives the respective times for the money to grow to $2100.

Question 4

Solve the problem.
-There are currently 57 million cars in a certain country, increasing exponentially by 0.8% annually. How many years will it take for this country to have 64 million cars? Round to the nearest year.

Accepted Answer

The number of cars grows according to the formula $N = N_0 \cdot (1 + r)^t$, where $N$ is the future amount, $N_0$ is the initial amount, $r$ is the rate of growth, and $t$ is time in years. Plugging in the given values: $64 = 57 \cdot (1 + 0.008)^t$. Solving for $t$ gives approximately 14 years.

Question 5

Solve the problem.
-Kimberly invested $3000 in her savings account for 7 years. When she withdrew it, she had $3779.58. Interest was compounded continuously. What was the interest rate on the account? Round to the nearest tenth of a percent.

Accepted Answer

The answer of Solve the problem.
-Kimberly invested $3000 in her...

Question 6

Solve the problem.
-The half-life of silicon-32 is 710 years. If 60 grams is present now, how much will be present in 300 years? (Round your answer to three decimal places.)

Accepted Answer

The answer of Solve the problem.
-The half-life of silicon-32 is...

Question 7

Solve the problem.
-A bacterial culture has an initial population of 10,000. If its population declines to 6000 in 8 hours, what will it be at the end of 10 hours? Assume that the population decreases according to the exponential model.

Accepted Answer

The exponential decay model is given by $P(t) = P_0e^{rt}$, where $P(t)$ is the population at time $t$, $P_0$ is the initial population, $r$ is the rate of decay, and $t$ is the time. Given $P_0 = 10000$ and $P(8) = 6000$, we can find $r$ by solving $6000 = 10000e^{8r}$. This gives $r = \frac{\ln(0.6)}{8}$. To find the population at $t = 10$, we substitute $r$ and $t = 10$ into the model: $P(10) = 10000e^{10\frac{\ln(0.6)}{8}}$, which calculates to approximately 5281.

Question 8

Solve the problem.
-How long will it take for an investment to triple in value if it earns 11.5% compounded continuously? Round your answer to three decimal places.

Accepted Answer

To find the time it takes for an investment to triple with continuous compounding, use the formula $ t = \frac{\ln(3)}{r} $, where $ r $ is the interest rate. Here, $ r = 0.115 $. Calculating gives $ t = \frac{\ln(3)}{0.115} \approx 9.553 $ years.

Question 9

Solve the equation. Express irrational answers in exact form and as a decimal rounded to 3 decimal places.
-ln x + ln (x + 6)= 4

Accepted Answer

The equation can be rewritten as ln(x(x+6)) = 4, leading to x(x+6) = e^4. Solving the quadratic equation x^2 + 6x - e^4 = 0 gives the solution x = (-6 + sqrt(36 + 4e^4))/2.

Question 10

Solve the problem.
-$$P ( t ) = \frac { 980 } { 1 + 20.78 e ^ { - 0.346 t } }$$ The logistic growth model represents the population of a bacterium in a culture tube after t hours. When will the amount of bacteria be 610?

Accepted Answer

The answer of Solve the problem.
-\[P ( t ) =...

Question 11

Solve the equation.
-$\log _ { 35 } ( x - 2 ) = 1 - \log _ { 35 } x$

Accepted Answer

To solve the equation, use the property of logarithms: $\log_b a - \log_b c = \log_b \frac{a}{c}$. Rewrite the equation as $\log_{35} (x - 2) + \log_{35} x = 1$, which simplifies to $\log_{35} \frac{x-2}{x} = 1$. This implies $\frac{x-2}{x} = 35^1 = 35$. Solving $\frac{x-2}{x} = 35$ gives $x = 7$.

Question 12

Solve the problem.
-How long will it take for $3000 to grow to $38,800 at an interest rate of 6.2% if the interest is compounded continuously? Round the number of years to the nearest hundredth.

Accepted Answer

The formula for continuous compounding is $A = Pe^{rt}$, 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), and $t$ is the time the money is invested for in years. Rearranging the formula to solve for $t$, we get $t = \frac{\ln(A/P)}{r}$. Substituting $A = 38800$, $P = 3000$, and $r = 0.062$ into the formula gives $t = \frac{\ln(38800/3000)}{0.062}$, which calculates to approximately 41.29 years.

Question 13

Solve the problem.
-Conservationists tagged 50 black-nosed rabbits in a national forest in 1990. In 1991, they tagged 100 black-nosed rabbits in the same range. If the rabbit population follows the exponential law, how many rabbits will be in the range 4 years from 1990?

Accepted Answer

The answer of Solve the problem.
-Conservationists tagged 50 black-nosed rabbits...

Question 14

Solve the equation.
-$$2 + \log _ { 3 } ( 2 x + 5 ) - \log _ { 3 } x = 4$$

Accepted Answer

The answer of Solve the equation.
-\[2 + \log _ {...

Question 15

Solve the equation.
-$$\log _ { 2 } ( x - 5 ) + \log _ { 2 } ( x - 11 ) = 4$$

Accepted Answer

The equation can be rewritten using the property of logarithms: $\log_2((x-5)(x-11)) = 4$. Converting to exponential form gives $(x-5)(x-11) = 2^4 = 16$. Solving the quadratic equation $x^2 - 16x + 55 = 0$ gives roots $x = 13$ and $x = 3$. However, $x = 3$ does not satisfy the domain of the original logarithmic expressions, as it makes $\log_2(x-11)$ undefined. Therefore, the only valid solution is $x = 13$.

Question 16

Solve the equation.
-$\log _ { 2 } ( x + 4 ) = 1 + \log _ { 2 } ( x - 3 )$

Accepted Answer

To solve the equation $\log_2(x + 4) = 1 + \log_2(x - 3)$, we first use the property of logarithms to combine the right side: $1 = \log_2(2)$ because $2^1 = 2$. This gives us $\log_2(x + 4) = \log_2(2) + \log_2(x - 3) = \log_2(2(x - 3))$. Since the bases are the same, we can set the arguments equal to each other: $x + 4 = 2(x - 3)$. Solving this equation, we get $x + 4 = 2x - 6$, which simplifies to $x = 10$. Therefore, the solution is $x = 10$.

Question 17

Solve the problem.
-$$P ( t ) = \frac { 170 } { 1 + 16 e ^ { - 0.188 t } }$$ territory after t years. When will the population be 80?

Accepted Answer

To find when the population will be 80, we set $P(t) = 80$ and solve for $t$. Substituting 80 for $P(t)$ gives us:$$80 = \frac{170}{1 + 16e^{-0.188t}}$$Multiplying both sides by the denominator and then dividing by 80 gives:$$1 + 16e^{-0.188t} = \frac{170}{80}$$$$1 + 16e^{-0.188t} = 2.125$$Subtracting 1 from both sides:$$16e^{-0.188t} = 1.125$$Dividing by 16:$$e^{-0.188t} = \frac{1.125}{16}$$$$e^{-0.188t} = 0.0703125$$Taking the natural logarithm of both sides to solve for $t$:$$-0.188t = \ln(0.0703125)$$$$t = \frac{\ln(0.0703125)}{-0.188}$$$$t \approx 14.12$$ years

Question 18

Solve the problem.
-The half-life of plutonium-234 is 9 hours. If 80 milligrams is present now, how much will be present in 6 days? (Round your answer to three decimal places.)

Accepted Answer

The answer of Solve the problem.
-The half-life of plutonium-234 is...

Question 19

Solve the equation.
-$$\frac { 1 } { 3 } \log _ { 2 } ( x + 6 ) = \log _ { 8 } ( 3 x )$$

Accepted Answer

By converting the logarithms to the same base and solving the equation, we find that $x = 3$ is the only solution that satisfies the equation.

Question 20

Solve the equation.
-$$\log _ { 2 } ( 3 x - 2 ) - \log _ { 2 } ( x - 5 ) = 4$$

Accepted Answer

Using the properties of logarithms, the equation can be rewritten as $\log_2\left(\frac{3x-2}{x-5}\right) = 4$. Solving for $x$ gives $x = 6$.

Quiz 5: Exponential and Logarithmic Functions and Equations

Solve the problem. -A fossilized leaf contains 12% of its normal amount of carbon 14. How old is the fossil (to the nearest year) ? Use 5600 years as the half-life of carbon 14.

Solve the problem. -How long will it take for an investment to double in value if it earns 4.25% compounded continuously? Round your answer to three decimal places.

Solve the problem. -If Emery has $1100 to invest at 6% per year compounded monthly, how long will it be before he has $2100? If the compounding is continuous, how long will it be? (Round your answers to three decimal places.)

Solve the problem. -There are currently 57 million cars in a certain country, increasing exponentially by 0.8% annually. How many years will it take for this country to have 64 million cars? Round to the nearest year.

Solve the problem. -Kimberly invested $3000 in her savings account for 7 years. When she withdrew it, she had $3779.58. Interest was compounded continuously. What was the interest rate on the account? Round to the nearest tenth of a percent.

Solve the problem. -The half-life of silicon-32 is 710 years. If 60 grams is present now, how much will be present in 300 years? (Round your answer to three decimal places.)

Solve the problem. -A bacterial culture has an initial population of 10,000. If its population declines to 6000 in 8 hours, what will it be at the end of 10 hours? Assume that the population decreases according to the exponential model.

Solve the problem. -How long will it take for an investment to triple in value if it earns 11.5% compounded continuously? Round your answer to three decimal places.

Solve the equation. Express irrational answers in exact form and as a decimal rounded to 3 decimal places. -ln x + ln (x + 6) = 4

Solve the problem. - $P ( t ) = \frac { 980 } { 1 + 20.78 e ^ { - 0.346 t } }$ The logistic growth model represents the population of a bacterium in a culture tube after t hours. When will the amount of bacteria be 610?

Solve the equation. - $\log _ { 35 } ( x - 2 ) = 1 - \log _ { 35 } x$

Solve the problem. -How long will it take for $3000 to grow to $38,800 at an interest rate of 6.2% if the interest is compounded continuously? Round the number of years to the nearest hundredth.

Solve the problem. -Conservationists tagged 50 black-nosed rabbits in a national forest in 1990. In 1991, they tagged 100 black-nosed rabbits in the same range. If the rabbit population follows the exponential law, how many rabbits will be in the range 4 years from 1990?

Solve the equation. - $2 + \log _ { 3 } ( 2 x + 5 ) - \log _ { 3 } x = 4$

Solve the equation. - $\log _ { 2 } ( x - 5 ) + \log _ { 2 } ( x - 11 ) = 4$

Solve the equation. - $\log _ { 2 } ( x + 4 ) = 1 + \log _ { 2 } ( x - 3 )$

Solve the problem. - $P ( t ) = \frac { 170 } { 1 + 16 e ^ { - 0.188 t } }$ territory after t years. When will the population be 80?

Solve the problem. -The half-life of plutonium-234 is 9 hours. If 80 milligrams is present now, how much will be present in 6 days? (Round your answer to three decimal places.)

Solve the equation. - $\frac { 1 } { 3 } \log _ { 2 } ( x + 6 ) = \log _ { 8 } ( 3 x )$

Solve the equation. - $\log _ { 2 } ( 3 x - 2 ) - \log _ { 2 } ( x - 5 ) = 4$