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Precalculus : Exponential and Logarithmic Functions

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Example Questions

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 #121 : Exponential And Logarithmic Functions

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=0

x=1

x=2

x=ln0

x=e

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 : Solve Exponential Equations

Solve an exponential equation.

Solve for .

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

Possible Answers:

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

x=ln15-6

x=3-ln3

x=6-ln2

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

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 : Solve Exponential Equations

Solve for .

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

Possible Answers:

x=e^2

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

x=e

$x=\ln(2e)$

x=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 #2 : 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: 1552 fish

2015: 523 fish

2010: 1046 fish

2015: 634 fish

2010: 634 fish

2015: 523 fish

2010: 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 #3 : Solve Exponential Equations

Solve for using properties of exponents.

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

Possible Answers:

x=3

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

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

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

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

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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Example Question #1 : Solve Exponential Equations

Solve: $6\mathrm{e}^{4x}-16=8$

Possible Answers:

$x=\ln\frac{1}{4}$

None of the other answers.

$x=\frac{\ln4}{4}$

$x=\ln4$

x=0

Correct answer:

$x=\frac{\ln4}{4}$

Explanation:

$6\mathrm{e}^{4x}-16=8$

Combine the constants:

$6\mathrm{e}^{4x}=24$

Isolate the exponential function by dividing:

$\mathrm{e}^{4x}=4$

Take the natural log of both sides:

$\ln(\mathrm{e}^{4x})=\ln(4)$

$4x=\ln(4)$

Finally isolate x:

$x=\frac{\ln(4)}{4}\hspace{1mm}or\hspace{1mm}.35$

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Example Question #2 : Solve Exponential Equations

Solve the equation for .

$2e^{2x}-7e^{x}+6=0$

Possible Answers:

$x=\pm ln3$

x=3,x=ln4

$x=ln2,x=ln(\frac{3}{2})$

$x=-ln(\frac{3}{4}),x=ln3$

x=ln(-2),x=ln7

Correct answer:

$x=ln2,x=ln(\frac{3}{2})$

Explanation:

$2e^{2x}-7e^{x}+6=0$

The key to this is that $e^{2x}=e^{x}\cdot e^{x}$ . From here, the equation can be factored as if it were $2x^{2}-7x+6=0$ .

$(2e^{x}-3)(e^{x}-2)=0$

$(2e^{x}-3)=0$ and $(e^{x}-2)=0$

$2e^{x}=3$ and $e^{x}=2$

$e^{x}=\frac{3}{2}$ and $e^{x}=2$

Now take the natural log (ln) of the two equations.

$ln(e^{x})=ln(\frac{3}{2})$ and $ln(e^{x})=ln2$

$x=ln\frac{3}{2}$ and x=ln2

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Example Question #1 : Graph Exponential Functions

Choose the description below that matches the equation:

y = 2^x

Possible Answers:

Exponential growth

Y-intercept at y = 0

Exponential decay

Y-intercept at y = 0

Exponential growth

Y-intercept at y=1

Exponential growth

Y-intercept at y=2

Exponential decay

Y-intercept at y = 1

Correct answer:

Exponential growth

Y-intercept at y=1

Explanation:

Exponential graphs can either decay or grow. This is based on the value of the base of the exponent. If the base is greater than , the graph will be growth. And, if the base is less than , then the graph will be decay. In this situation, our base is . Since this is greater than , we have a growth graph. Then, to determine the y-intercept we substitute x=0 . Thus, we get:

y = 2^0 = 1 for the y-intercept.

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Precalculus : Exponential and Logarithmic Functions

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #1214 : Pre Calculus

Example Question #121 : Exponential And Logarithmic Functions

Example Question #1 : Solve Exponential Equations

Example Question #2 : Solve Exponential Equations

Example Question #3 : Solve Exponential Equations

Example Question #2 : Solve Exponential Equations

Example Question #3 : Solve Exponential Equations

Example Question #1 : Solve Exponential Equations

Example Question #2 : Solve Exponential Equations

Example Question #1 : Graph Exponential Functions

All Precalculus Resources

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