Exponential Functions - GRE Subject Test: Math

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Question

Suppose a blood cell increases proportionally to the present amount. If there were blood cells to begin with, and blood cells are present after hours, what is the growth constant?

Answer

The population size after some time is given by:

where is the initial population.

At the start, there were 30 blood cells.

Substitute this value into the given formula.

After 2 hours, 45 blood cells were present. Write this in mathematical form.

Substitute this into , and solve for .

$\frac{3}{2}= e^2^k$

$ln\left(\frac{3}{2}\right)=ln(e^2^k)$

$2k=ln\left(\frac{3}{2}\right)$

$k=\frac{1}{2}ln\left(\frac{3}{2}\right)$

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Question

Suppose a population of bacteria increases from to in . What is the constant of growth?

Answer

The equation for population growth is given by $P=Ae^{kt}$ . is the population, is the intial value, is time, and is the growth constant. We can plug in the values we know at time and solve for .

$100=Ae^{0k}=A$

Now that we solved for , we can plug in what we know for time and solve for .

$1000=100e^{30k}$

$e^{30k}=10$

$30k=\ln(10)$

$k=\frac{\ln(10)}{30}$

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Question

A population of deer grew from 50 to 200 in 7 years. What is the growth constant for this population?

Answer

The equation for population growth is given by $P=Ae^{kt}$ . P is the population, is the intial value, is time, and is the growth constant. We can plug in the values we know at time and solve for .

$50=Ae^{0k}=A$

Now that we have solved for we can solve for at

$200=50e^{7k}$

$e^{7k}=4$

$7k=\ln(4)$

$k=\frac{\ln(4)}{7}$

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Question

A population of mice has 200 mice. After 6 weeks, there are 1600 mice in the population. What is the constant of growth?

Answer

The equation for population growth is given by $P=Ae^{kt}$ . is the population, is the intial value, is time, and is the growth constant. We can plug in the values we know at time and solve for .

$200=Ae^{0k}=A$

Now that we have we can solve for at .

$1600=200e^{6k}$

$e^{6k}=8$

$6k=\ln(8)$

$k=\frac{\ln(8)}{6}$

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Question

The rate of growth of the duck population in Wingfield is proportional to the population. The population increased by 15 percent between 2001 and 2008. What is the constant of proportionality?

Answer

We're told that the rate of growth of the population is proportional to the population itself, meaning that this problem deals with exponential growth/decay. The population can be modeled thusly:

$p(t)=p_0e^{kt}$

Where is an initial population value, and is the constant of proportionality.

Since the population increased by 15 percent between 2001 and 2008, we can solve for this constant of proportionality:

$(1+.15)y_0=y_0e^{k(2008-2001)}$

$1.15=e^{7k}$

$k=\frac{ln(1.15)}{7}=0.01997$

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Question

The rate of growth of the bacteria in an agar dish is proportional to the population. The population increased by 150 percent between 1:15 and 2:30. What is the constant of proportionality?

Answer

We're told that the rate of growth of the population is proportional to the population itself, meaning that this problem deals with exponential growth/decay. The population can be modeled thusly:

$p(t)=p_0e^{kt}$

Where is an initial population value, and is the constant of proportionality.

Since the population increased by 150 percent between 1:15 and 2:30, we can solve for this constant of proportionality:

$(1+1.50)y_0=y_0e^{k(2:30 - 1:15)}$

Dealing in minutes:

$2.50=e^{75k}$

$k=\frac{ln(2.50)}{75}=0.012$

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Question

The rate of growth of the Martian Transgalactic Constituency is proportional to the population. The population increased by 23 percent between 2530 and 2534 AD. What is the constant of proportionality?

Answer

We're told that the rate of growth of the population is proportional to the population itself, meaning that this problem deals with exponential growth/decay. The population can be modeled thusly:

$p(t)=p_0e^{kt}$

Where is an initial population value, and is the constant of proportionality.

Since the population increased by 23 percent between 2530 and 2534 AD, we can solve for this constant of proportionality:

$(1+0.23)y_0=y_0e^{k(2534-2530)}$

$1.23=e^{4k}$

$k=\frac{ln(1.23)}{4}=0.05$

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Question

The rate of decrease of the dwindling wolf population of Zion National Park is proportional to the population. The population decreased by 7 percent between 2009 and 2011. What is the constant of proportionality?

Answer

We're told that the rate of growth of the population is proportional to the population itself, meaning that this problem deals with exponential growth/decay. The population can be modeled thusly:

$p(t)=p_0e^{kt}$

Where is an initial population value, and is the constant of proportionality.

Since the population decreased by 7 percent between 2009 and 2011, we can solve for this constant of proportionality:

$(1-0.07)y_0=y_0e^{k(2011-2009)}$

$0.93=e^{2k}$

$k=\frac{ln(0.93)}{2}=-0.036$

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Question

The rate of decrease of the panda population is proportional to the population. The population decreased by 12 percent between 1990 and 2001. What is the constant of proportionality?

Answer

We're told that the rate of growth of the population is proportional to the population itself, meaning that this problem deals with exponential growth/decay. The population can be modeled thusly:

$p(t)=p_0e^{kt}$

Where is an initial population value, and is the constant of proportionality.

Since the population decreased by 12 percent between 1990 and 2001, we can solve for this constant of proportionality:

$(1-0.12)y_0=y_0e^{k(2001-1990)}$

$0.88=e^{11k}$

$k=\frac{ln(0.88)}{11}=-0.012$

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Question

The rate of growth of the salmon population of Yuba is proportional to the population. The population increased by 21 percent over the course of seven years. What is the constant of proportionality?

Answer

We're told that the rate of growth of the population is proportional to the population itself, meaning that this problem deals with exponential growth/decay. The population can be modeled thusly:

$p(t)=p_0e^{kt}$

Where is an initial population value, and is the constant of proportionality.

Since the population increased by 21 percent over the course of seven years, we can solve for this constant of proportionality:

$(1+0.21)y_0=y_0e^{7k}$

$1.21=e^{7k}$

$k=\frac{ln(1.21)}{7}=0.027$

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Question

The rate of decrease of the number of concert attendees to former teen heartthrob Justice Beaver is proportional to the population. The population decreased by 34 percent between 2013 and 2015. What is the constant of proportionality?

Answer

We're told that the rate of growth of the population is proportional to the population itself, meaning that this problem deals with exponential growth/decay. The population can be modeled thusly:

$p(t)=p_0e^{kt}$

Where is an initial population value, and is the constant of proportionality.

Since the population decreased by 34 percent between 2013 and 2015, we can solve for this constant of proportionality:

$(1-0.34)y_0=y_0e^{k(2015-2013)}$

$0.66=e^{2k}$

$k=\frac{ln(0.66)}{2}=-0.21$

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Question

The rate of growth of the Land of Battlecraft players is proportional to the population. The population increased by 72 percent between February and October of 2015. What is the constant of proportionality?

Answer

We're told that the rate of growth of the population is proportional to the population itself, meaning that this problem deals with exponential growth/decay. The population can be modeled thusly:

$p(t)=p_0e^{kt}$

Where is an initial population value, and is the constant of proportionality.

Since the population increased by 72 percent between February and October, we can solve for this constant of proportionality. It'll help to represent the months by their number in the year:

$(1+0.72)y_0=y_0e^{k(10-2)}$

$1.72=e^{8k}$

$k=\frac{ln(1.72)}{8}=0.068$

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Question

The rate of decrease of the gluten-eating demographic of the US is proportional to the population. The population decreased by 8 percent between 2014 and 2015. What is the constant of proportionality?

Answer

We're told that the rate of growth of the population is proportional to the population itself, meaning that this problem deals with exponential growth/decay. The population can be modeled thusly:

$p(t)=p_0e^{kt}$

Where is an initial population value, and is the constant of proportionality.

Since the population decreased by 8 percent between 2014 and 2015, we can solve for this constant of proportionality:

$(1-0.08)y_0=y_0e^{k(2015-2014)}$

$0.92=e^{k}$

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Question

Bob invests in a bank that compounds interest continuously at a rate of . How much money will Bob have in his account after years? (Round answer to decimal places.)

Answer

Step 1: Recall the formula for continuously compounded interest

The formula is: $A=Pe^{rt}$ , where:

is the Final balance after years.

is the original investment balance.

is the exponential function

is the interest rate, usually written as a decimal

is the time, usually in years

Step 2: Plug in all the information that we have into the formula

$A=($10,000)e^{0.12(3)}$

Simplify:

$A=($10,000)e^{.36}$

Step 3: Evaluate.

$e^{.36}\approx1.433329$

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Question

Find one possible value of , given the following equation:

$343=7^{15t-12}$

Answer

We begin with the following:

$343=7^{15t-12}$

This can be rewritten as

$7^3=7^{15t-12}$

Recall that if you have two exponents with equal bases, you can simply set the exponents equal to eachother. Do so to get the following:

Solve this to get t.

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Question

Solve for .

$4^{2x}=16^{2x-6}$

Answer

We need to make the bases equal before attempting to solve for . Since we can rewrite our equation as

$4^{2x}=4^{2^{2x-6}}$

Remember: the exponent rule $(x^n)^m=x^{n\cdot m}$

$4^{2x}=4^{2(2x-6)}$

Now that our bases are equal, we can set the exponents equal to each other and solve for .

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Question

Solve for .

$7^{2y}-3^{y+2}=0$

Answer

The first step is to make sure we don't have a zero on one side which we can easily take care of:

$7^{2y}=3^{y+2}$

Now we can take the logarithm of both sides using natural log:

$\ln 7^{2y}=\ln 3^{y+2}$

Note: we can apply the Power Rule here $\ln (x^y)=y\ln (x)$

$2y \ln 7=(y+2)\ln3$

$2y \ln 7=y\ln3+2\ln3$

$2y \ln 7-y\ln3=2\ln3$

$y(2 \ln 7-\ln3)=2\ln3$

$y=\frac{2\ln3}{2 \ln 7-\ln3}$

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Question

Solve for .

$3e^{(4x-5)}=24$

Answer

Before beginning to solve for , we need to have a coefficient of :

$e^{(4x-5)}=8$

Now we can take the natural log of both sides:

$\ln e^{(4x-5)}=\ln8$

Note: $\ln (e)=1$

$4x-5=\ln8$

$4x=\ln8+5$

$x=\frac{\ln8+5}{4}$

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Question

$6^{3}\times6^{n} =6^{12}$

Answer

$6^{3} \times 6^{n} =6^{12}$

Since the base is for both, then:

When the base is the same, and you are multiplying, the exponents are added.

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Question

$2^{x} = 40$

Answer

$2^{x} = 40$

To solve, use the natural log.

$ln (2^{x}) = ln (40)$

To isolate the variable, divide both sides by .

$x = \frac{\ln (40)}{\ln (2)}$

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