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College Algebra : Exponential and Logarithmic Functions

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Example Questions

Example Question #2 : Solving Logarithms

Solve the equation:

$\log{_{4}}{x^{2}}+\log{_{4}}{3}=1$

Possible Answers:

$\frac{\sqrt{6}}{3}, -\frac{\sqrt{6}}{3}$

$\frac{\sqrt{3}}{3}, -\frac{\sqrt{3}}{3}$

$\frac{\sqrt{3}}{2}, -\frac{\sqrt{3}}{2}$

1,-1

$\frac{2\sqrt{3}}{3}, -\frac{2\sqrt{3}}{3}$

Correct answer:

$\frac{2\sqrt{3}}{3}, -\frac{2\sqrt{3}}{3}$

Explanation:

$\log{_{4}}{x^{2}}+\log{_{4}}{3}=1$

$\Rightarrow \log{_{4}}{x^{2}}+\log{_{4}}{3}=\log{_{4}}{4}$

$\Rightarrow log{_{4}}{(x^{2}\times 3)}=\log{_{4}}{4}$

$\Rightarrow 3x^{2}=4$

$\Rightarrow x^{2}=\frac{4}{3}$

$\Rightarrow x=\frac{2\sqrt{3}}{3}, x=-\frac{2\sqrt{3}}{3}$

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Example Question #3 : Solving Logarithms

Solve for .

$log_{9}^{ }(2x+1)=2$

Possible Answers:

x=40

$x=\frac{17}{2}$

$x=\frac{81}{2}$

x=20

x=41

Correct answer:

x=40

Explanation:

Rewrite in exponential form:

$9^{2}=2x+1$

Solve for x:

81=2x+1

x=40

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Example Question #4 : Solving Logarithms

Solve the following equation:

ln(3x+8)=0

Possible Answers:

$-\frac{8}{3}$

$\frac{8}{7}$

$-\frac{7}{3}$

$\frac{1}{3}$

Correct answer:

$-\frac{7}{3}$

Explanation:

For this problem it is helpful to remember that,

ln(3x+8) = 0 is equivalent to ln(3x+8) = ln(1) because ln(1)=0

Therefore we can set what is inside of the parentheses equal to each other and solve for as follows:

3x+8=1

3x=-7

$x=-\frac{7}{3}$

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Example Question #11 : Solving Logarithms

Solve this logarithmic equation: $\log_{10}(5x-2)=\log_{10}(x+6)$

Possible Answers:

x=1

x=.5

x=2

None of the other answers.

x=4

Correct answer:

x=2

Explanation:

To solve this problem you must be familiar with the one-to-one logarithmic property.

$\log_{10}x=\log_{10}y$ if and only if x=y. This allows us to eliminate to logarithmic functions assuming they have the same base.

$\log_{10}(5x-2)=\log_{10}(x+6)$

one-to-one property:

(5x-2)=(x+6)

isolate x's to one side:

4x-2=6

move constant:

4x=8

x=2

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Example Question #2 : Solving Logarithmic Functions

Solve the equation:

$5e^{3x+4}-10=0$

Possible Answers:

$x=\frac{-4+ln2}{3}$

$x=-\frac{4}{3}+ln10$

No solution exists

x=-4+ln2

$x=\frac{ln6}{3}$

Correct answer:

$x=\frac{-4+ln2}{3}$

Explanation:

Get all the terms with e on one side of the equation and constants on the other.

$5e^{3x+4}-10=0$

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

Apply the logarithmic function to both sides of the equation.

$ln(e^{3x+4})=ln2$

3x+4=ln2

3x=-4+ln2

$x=\frac{-4+ln2}{3}$

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Example Question #12 : Solving Logarithms

Solve the equation:

2log(x)+log(5)=3

Possible Answers:

$x=\pm 10\sqrt{2}$

$x=\pm \sqrt{995}$

$x=\pm6\sqrt{2}$

$x=\pm 12$

$x=\pm3\sqrt{5}$

Correct answer:

$x=\pm 10\sqrt{2}$

Explanation:

Recall the rules of logs to solve this problem.

First, when there is a coefficient in front of log, this is the same as log with the inside term raised to the outside coefficient.

2log(x)+log(5)=3

$log(x^{2})+log(5)=3$

Also, when logs of the same base are added together, that is the same as the two inside terms multiplied together.

In mathematical terms:

$log(a)+log(b)=log(a\cdot b)$

Thus our equation becomes,

$log(5x^{2})=3$

To simplify further use the rule,

$log_b(x)\rightarrow b^{log_b(x)}=x$

$10^{log(5x^{2})}=10^{3}$

$5x^{2}=1000$

$x^{2}=200$

$x=\pm10\sqrt{2}$ .

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Example Question #3 : Solving Logarithmic Functions

Solve the following equation for t

log_t(625)=4

Possible Answers:

t=25

t=3

t=5

t=125

Correct answer:

t=5

Explanation:

Solve the following equation for t

log_t(625)=4

We can solve this equation by rewriting it as an exponential equation:

t^4=625

Next, take the fourth root to get:

$t=\sqrt[4]{625}=5$

We can check our work vie the following:

5^4=625

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Example Question #4 : Solving Logarithmic Functions

Solve for ,

$\ln(x^2 +1)=3$

Possible Answers:

$x = \pm \sqrt{e^3-1}$

$x = \pm 2$

$x = \pm\frac{7\sqrt{e}}{3}$

It is impossible to isolate .

$x = \pm \frac{3}{2}\sqrt{1+e}$

Correct answer:

$x = \pm \sqrt{e^3-1}$

Explanation:

$\ln(x^2+1)=3$

The first step step is to carry out the inverse operation of the natural logarithm,

$e^{\ln(x^2+1)}=e^3$

We can use the property $e^{ln(u)}=u$ to simplify the left side of the equation to obtain,

x^2+1 = e^3

Solve for ,

$x = \pm \sqrt{e^3-1}$

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Example Question #4 : Solving Logarithmic Functions

Solve this equation: $e^{2x}-3e^x+2=0$

Possible Answers:

x=ln 2

x=0

x=ln2

x=1

None of these.

No solution.

x=ln3

x=1

Correct answer:

x=ln 2

x=0

Explanation:

$e^{2x}-3e^x+2=0$

Note that this equation is a quadratic because $e^{2x}=(e^x)^2$

Factor:

(e^x-2)(e^x-1)

Set each factor equal to 0 and solve:

$e^x-2=0\rightarrow e^x=2 \rightarrow \boldsymbol{x=ln\hspace{1mm} 2}$

$e^x-1=0\rightarrow e^x=1 \rightarrow \boldsymbol{x=0}$

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Example Question #55 : Exponential And Logarithmic Functions

Solve this logarithmic equation: ln(2x-2)=5

Possible Answers:

x=75.2

x=27.5

x=72.5

x=52.7

None of these.

Correct answer:

x=75.2

Explanation:

ln(2x-2)=5

Exponentiate:

$e^{ln2x-2}=e^5$

Add two to both sides:

$2x-2=e^5\rightarrow 2x=e^5+2$

Divide both sides by 2:

$x=\frac{e^5}{2}+1=\boldsymbol{75.2}$ (rounded answer)

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College Algebra : Exponential and Logarithmic Functions

Study concepts, example questions & explanations for College Algebra

All College Algebra Resources

Example Questions

Example Question #2 : Solving Logarithms

Example Question #3 : Solving Logarithms

Example Question #4 : Solving Logarithms

Example Question #11 : Solving Logarithms

Example Question #2 : Solving Logarithmic Functions

Example Question #12 : Solving Logarithms

Example Question #3 : Solving Logarithmic Functions

Example Question #4 : Solving Logarithmic Functions

Example Question #4 : Solving Logarithmic Functions

Example Question #55 : Exponential And Logarithmic Functions

All College Algebra Resources

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