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GRE Subject Test: Math : Exponential Functions

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Example Question #21 : Solving Exponential Equations

Solve for :

$2^{x+1}=8^4$

Possible Answers:

x=-7

x=4

x=-11

x=11

Correct answer:

x=11

Explanation:

Step 1: Rewrite the right side as a power of :

$8^4=8\cdot8\cdot8\cdot8$

8=2^3

$8^4=2^3\cdot2^3\cdot2^3\cdot2^3$

$8^4=2^{3+3+3+3}=2^{3\cdot 4}=2^{12}$

Step 2: Rewrite the original equation:

$2^{x+1}=2^{12}$

Step 3: Since the bases are equal, I can set the exponents equal.

So, $x+1=12\implies x=11$

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Example Question #341 : Gre Subject Test: Math

Solve for :

$2^{x+3}=8^4$

Possible Answers:

Correct answer:

Explanation:

Rewrite the right hand side such that it has the same base as the expression on the right hand side.

$8^4=\big(2^3\big)^{4}$
Apply the rule of stacked exponents. We multiply the exponents together and keep the same base:

$\large \big(2^3\big)^4=2^{3\cdot4}=2^{12}$

Now that we have rewrote the right hand side in the same way as the left, we can equate them:

$\large 2^{x+3}=2^{12}$

Now that we have the same bases, we can equal the exponents and solve for x...

$\large x+3=12 \implies x=9$

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Example Question #23 : Solving Exponential Equations

Find the value that makes the statement $(36)^{x+2}=6$ true.

Possible Answers:

$\frac {1}{2}$

$\frac {3}{2}$

$-\frac {3}{2}$

Correct answer:

$-\frac {3}{2}$

Explanation:

Step 1: Re-write the left side so we have the same base as the right side...

$(36)^{x+2}=6\implies (6^2)^{x+2}=6^1$

Step 2: Multiply the stacked exponents on the left hand side...

We will apply the rule: $(a^b)^c=a^{b\cdot c}$

So, $(6^2)^{x+2}=6^{2(x+2)}=6^{2x+4}$

Step 3: We can now set the exponents equal to each other, since we have the same base.

2x+4=1

2x=1-4

2x=-3

$\frac {{\color{Blue} 2}x}{{\color{Blue} 2}}=-\frac {{\color{Red} 3}}{{\color{Blue} 2}}$

$x=-\frac {3}{2}$

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Example Question #21 : Solving Exponential Equations

Find one possible value of , given the following equation:

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

Possible Answers:

Cannot be determined from the information given.

t=1

t=5

t=0

t=-3

Correct answer:

t=1

Explanation:

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:

3=15t-12

Solve this to get t.

15=15t

t=1

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Example Question #1 : Exponential Growth Applications

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?

Possible Answers:

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

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

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

$\frac{2}{3}$

$\frac{3}{2}$

Correct answer:

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

Explanation:

The population size P(t) after some time is given by:

P(t)= P_ie^k^t

where P_i is the initial population.

At the start, there were 30 blood cells.

P_i=P(0)=30

Substitute this value into the given formula.

P(t)= 30e^k^t

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

P(2)=45

Substitute this into P(t)= 30e^k^t , and solve for .

45= 30e^k^(^2^)

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

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

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

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

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Example Question #2 : Exponential Growth Applications

Suppose a population of bacteria increases from 100 to 1000 in $30\: min$ . What is the constant of growth?

Possible Answers:

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

$e^{30}$

$\ln(10)$

None of these

Correct answer:

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

Explanation:

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 t=0 and solve for .

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

Now that we solved for , we can plug in what we know for time t=30 and solve for .

$1000=100e^{30k}$

$e^{30k}=10$

$30k=\ln(10)$

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

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Example Question #3 : Exponential Growth Applications

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

Possible Answers:

$\frac{4}{7}$

$\ln(4)$

$\ln(7)$

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

None of these

Correct answer:

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

Explanation:

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 t=0 and solve for .

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

Now that we have solved for we can solve for at t=7

$200=50e^{7k}$

$e^{7k}=4$

$7k=\ln(4)$

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

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Example Question #4 : Exponential Growth Applications

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

Possible Answers:

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

$\ln(8)$

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

$\frac{\ln(6)}{6}$

Correct answer:

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

Explanation:

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 t=0 and solve for .

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

Now that we have we can solve for at t=6 .

$1600=200e^{6k}$

$e^{6k}=8$

$6k=\ln(8)$

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

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Example Question #1 : How To Find Constant Of Proportionality Of Rate

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?

Possible Answers:

-0.036

-0.044

-0.009

-0.011

0.130

Correct answer:

-0.036

Explanation:

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 p_0 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:

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

$0.93=e^{2k}$

2k=ln(0.93)

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

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Example Question #6 : Exponential Growth Applications

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?

Possible Answers:

-0.37

1.04

0.12

0.05

0.07

Correct answer:

0.05

Explanation:

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 p_0 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:

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

$1.23=e^{4k}$

4k=ln(1.23)

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

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GRE Subject Test: Math : Exponential Functions

Study concepts, example questions & explanations for GRE Subject Test: Math

All GRE Subject Test: Math Resources

Example Questions

Example Question #21 : Solving Exponential Equations

Example Question #341 : Gre Subject Test: Math

Example Question #23 : Solving Exponential Equations

Example Question #21 : Solving Exponential Equations

Example Question #1 : Exponential Growth Applications

Example Question #2 : Exponential Growth Applications

Example Question #3 : Exponential Growth Applications

Example Question #4 : Exponential Growth Applications

Example Question #1 : How To Find Constant Of Proportionality Of Rate

Example Question #6 : Exponential Growth Applications

All GRE Subject Test: Math Resources

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