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Precalculus : Pre-Calculus

Study concepts, example questions & explanations for Precalculus

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12 Diagnostic Tests 380 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #1211 : Pre Calculus

Evaluate the logarithm

$\small \log_4 (\frac{1}{4}) \,$

Possible Answers:

$\small \frac{1}{16}$

$\small -\frac{3}4{}$

Correct answer:

Explanation:

Given the property

$\small \log_a a^b = b$

and because

$\small 4^{-1} = \frac{1}{4}$

we obtain that

$\small \small \log_4 (\frac{1}{4}) =-1$

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Example Question #1212 : Pre Calculus

Evaluate to four decimal places: $\log _{5} 7$

Possible Answers:

0.1690

0.8271

0.1461

1.2091

4.8928

Correct answer:

1.2091

Explanation:

Using a calculator, evaluate $\frac{\log7}{\log5} \approx 1.2091$

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Example Question #1213 : Pre Calculus

Evaluate to 4 decimal places: $\log _{4 } 80$

Possible Answers:

0.3164

3.1610

1.3010

48.1648

0.4758

Correct answer:

3.1610

Explanation:

To evaluate, use a calculator to find $\frac{\log80}{\log4} \approx 3.1610$

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Example Question #1214 : Pre Calculus

Evaluate the following:

$\log_{2}4-\log_{3}27$

Possible Answers:

Correct answer:

Explanation:

To solve, simply find the matching base and use properties of logs to simplify.

Some properties of logs include the following:

1) $log_bx=y\Rightarrow b^y=x$

2) $logx^y\Rightarrow ylogx$

3) log_xx=1

Thus,

$\\ \log_{2}4-\log_{3}27\\ \\ =\log_{2}2^2-\log_{3}3^3\\ \\=2\log_{2}2-3\log_{3}3\\ \\=2-3\\ \\=-1$

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Example Question #1214 : Pre Calculus

Evaluate the following:

$2\log2+\log\frac{5}{2}$

Possible Answers:

$\frac{5}{2}$

Correct answer:

Explanation:

To solve, remember the following rules for logarithms.

${x}\log{a}=\log{a^x}$

$\log{a}+\log{b}=\log{ab}$

Thus,

$2\log2+\log\frac{5}{2}=\log{2^2}+\log{\frac{5}{2}}=\log{\frac{4*5}{2}}=\log{10}=1$

Remember, if a base isn't specified, it is 10.

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Example Question #1 : Solve Exponential Equations

Solve an equation involving exponents and logarithms.

Solve for .

$e^{-3}e^{2}lne^{e}=e^{x}$

Possible Answers:

x=e

x=1

x=2

x=ln0

x=0

Correct answer:

x=0

Explanation:

First, simplify the left side of the equation using the additive rule for exponents.

$e^{-3}e^{2}lne^{e}=(e^{(-3+2)})lne^{e}$ .

Our equation now becomes:

$e^{-1}lne^e=e^x$

$\frac{1}{e}lne^e=e^x$

$lne^e=e^x \cdot e$

$e=e^{x+1}$

Equating we set the exponents equal to eachother and solve.

1=x+1

Thus, x=0.

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Example Question #2 : Exponential Functions

Solve an exponential equation.

Solve for .

$e^{3x+2}=\frac{3e^{5x}}{e^{4}}$

Possible Answers:

$x=\frac{6-ln3}{2}$

x=3-ln3

$x=\frac{6-5ln3}{2}$

x=6-ln2

x=ln15-6

Correct answer:

$x=\frac{6-ln3}{2}$

Explanation:

First, use the additive property of exponents to simplify the right side of the equation.

$\frac{3e^{5x}}{e^{4}}=3e^{5x-4}$ .

Thus,

$e^{3x+2}=3e^{5x-4}$ .

Now, take the natural log of both sides

$ln(e^{3x+2})=ln(3e^{5x-4})$ .

Use the multiplicative property of logarithms to expand the left side to get

$ln(e^{3x+2})=ln3+lne^{5x-4}$

Now, apply the logarithms to the exponents

3x+2=ln3 +5x-4 .

Rearrange to get the x-terms on one side

2+4-ln3=5x-3x

6-ln3=2x .

Finally, divide the 2 on both sides

$x=\frac{6-ln3}{2}$ .

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Example Question #3 : Exponential Functions

Solve for .

$2e^5e^{-3}\ln(e^x)=2x^2$

Possible Answers:

x=e

$x=2\ln(e^2)$

x=2e

x=e^2

$x=\ln(2e)$

Correct answer:

x=e^2

Explanation:

First, let's begin by simplifying the left hand side.

$e^5e^{-3}$ becomes e^2 and $\ln(e^x)$ becomes . Remember that $\ln(e)=1$ , and the in that expression can come out to the front, as in $x\cdot \ln(e)$ .

Now, our expression is

2e^2x = 2x^2

From this, we can cancel out the 2's and an x from both sides.

Thus our answer becomes:

x=e^2 .

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Example Question #4 : Solve Exponential Equations

The population of fish in a pond is modeled by the exponential function

$y=1046e^{-0.1t}$ , where is the population of fish and is the number of years since January 2010.

Determine the population of fish in January 2010 and January 2015.

Possible Answers:

2010: 634 fish

2015: 523 fish

2010: 1046 fish

2015: 634 fish

2010: fish

2015: 523 fish

2010: 1552 fish

2015: 523 fish

Correct answer:

2010: 1046 fish

2015: 634 fish

Explanation:

In 2010, t=0 in our equation because we have had no years past 2010. Plugging that in to the model equation and solving:

$y=1046e^{-0.1(0)} = 1046$ , since anything raised to the power of zero becomes . So the population of fish in 2010 is 1046 fish.

In 2015, t=5 because 5 years have passed since 2010. Plugging that into our equation and solving gives us

$y=1046e^{-0.1(5)} \approx 634$

So the population of fish in 2015 is 634 fish. This is an example of exponential decay since the function is decreasing.

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Example Question #5 : Solve Exponential Equations

Solve for using properties of exponents.

$3^{4x}=27^7$

Possible Answers:

$x=\frac{4}{21}$

$x=\frac{21}{4}$

x=3

$x=\frac{7}{4}$

$x=\frac{12}{7}$

Correct answer:

$x=\frac{21}{4}$

Explanation:

Since 27=3^3 , the equation simplifies to $3^{4x}=(3^3)^7=3^{21}$ .

Since the bases are equal, we can then set the exponents equal to each other.

Solving for x in this simple equation gives the correct answer.

$4x=21 \implies x=\frac{21}{4}$

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Precalculus : Pre-Calculus

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #1211 : Pre Calculus

Example Question #1212 : Pre Calculus

Example Question #1213 : Pre Calculus

Example Question #1214 : Pre Calculus

Example Question #1214 : Pre Calculus

Example Question #1 : Solve Exponential Equations

Example Question #2 : Exponential Functions

Example Question #3 : Exponential Functions

Example Question #4 : Solve Exponential Equations

Example Question #5 : Solve Exponential Equations

All Precalculus Resources

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