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Algebra II : Algebra II

Study concepts, example questions & explanations for Algebra II

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Example Questions

Example Question #31 : Solving And Graphing Logarithms

Solve the following equation:

log_2 (-x)=8

Possible Answers:

x=-256

x=-3

x=-16

x=3

x=256

Correct answer:

x=-256

Explanation:

Recall the definition of a logarithm. For a logarithm in any base ,

log_bx=c

denotes

b^c=x .

In other words, the value of a logarithm is simply an exponent, and it is defined at whenever the base can be raised to a sufficient power to yield . Hence, the equation

log_2(-x)=8

may be rewritten as

2^8=-x .

Raising to the th power yields 256 , and so

x=-256 .

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Example Question #32 : Solving Logarithms

Solve the following equation:

e^2^l^n^x=4

Possible Answers:

x=8,-8

x=2,-2

x=2

x=8

x=16

Correct answer:

x=2

Explanation:

According to the rule for multiplying logarithms, alogx=logx^a for any constant value . In other words, an exponent on the quantity inside a logarithm can be moved outside the logarithm as a multiplier, and vice versa. Hence,

$e^2^l^n^x=e^l^n^{x^2} =4$ .

Since the natural logarithm has as its base, the definition of logarithms applies in this case: b^l^o^g^x=x whenever the base of logx is equal to .

$e^l^n^{x^2}=x^2=4$ .

This implies that x=2 and x=-2 . However, substituting the solution x=-2 into the original equation yields the following expression:

e^2^l^n^(^-^2^)=4

which is not defined, since logarithms are undefined for any value of $x\leq 0$ .

Hence, we eliminate the extraneous solution x=-2 and submit that the correct solution to this equation is x=2 .

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Example Question #33 : Solving Logarithms

Solve for : log_e(2x) = -6

Possible Answers:

$\frac{1}{2e^6}$

$e^{12}$

$\frac{2}{e^6}$

$\frac{1}{2e^{12}}$

$\frac{2}{e^{12}}$

Correct answer:

$\frac{1}{2e^6}$

Explanation:

Use the following interchangeable properties to solve for the unknown variable:

log_a(b)= c

b=a^c

For the equation log_e(2x) = -6 , this would be equivalent to: $2x= e^{-6}$

Divide by two on both sides and convert the negative exponent to a fraction.

$\frac{2x}{2}= \frac{e^{-6}}{2}$

$x=\frac{1}{2}\cdot \frac{1}{e^6} = \frac{1}{2e^6}$

The answer is: $\frac{1}{2e^6}$

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Example Question #34 : Solving Logarithms

Solve the log: $log_{3}(9^5)$

Possible Answers:

125

Correct answer:

Explanation:

In order to simplify this log, we will need to convert the nine to base three.

9=3^2

Rewrite the term inside the log.

$log_{3}(9^5)=log_{3}(3^{2\times5})$

Rewrite this as a fraction.

$\frac{log(3^{2\times5})}{log(3)} = \frac{log(3^{10})}{log(3^1)}$

According to log rules, we can pull the power of the inner quantity inside the log out as the coefficient.

$\frac{10}{1}= 10$

The answer is:

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Example Question #38 : Solving Logarithms

Evaluate: $log_{100}10$

Possible Answers:

$\frac{1}{2}$

$\sqrt2$

$\sqrt e$

$\frac{1}{4}$

e^2

Correct answer:

$\frac{1}{2}$

Explanation:

The expression can be rewritten as a fraction.

$log_{100}10 = \frac{log(10)}{log(100)}$

The log based 100 can be rewritten as ten squared.

$\frac{log(10)}{log(100)} = \frac{log(10)}{log(10^2)} = \frac{log(10)}{2log(10)} = \frac{1}{2}$

The answer is: $\frac{1}{2}$

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Example Question #35 : Solving Logarithms

Solve the logs: log(2)+log(4)+log(6)-log(8)

Possible Answers:

3log(2)+2log(3)

log(4)

log(2)+log(3)

log(5)

3log(2)

Correct answer:

log(2)+log(3)

Explanation:

For logs that are added, the inner quantities can be multiplied with each other to be combined as a single log.

For logs that are subtracted, the inner quantities would each be divided instead.

Rewrite the expression:

$log(2)+log(4)+log(6)-log(8) =log( \frac{2(4)(6)}{8})$

= log(6)= log(2)+log(3)

The answer is: log(2)+log(3)

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Example Question #36 : Solving Logarithms

Evaluate: $log(\frac{x}{10})= 2$

Possible Answers:

$\frac{1}{100}$

1000

$\frac{1}{1000}$

100

$\frac{1}{200}$

Correct answer:

1000

Explanation:

Recall that log has a base of ten.

The equation can be rewritten as:

$log_{10}(\frac{x}{10})= 2$

$log_{10}(x)-log_{10}(10) = 2$

$log_{10}(x)-1= 2$

Add 1 on both sides.

$log_{10}(x)-1+1= 2+1$

$log_{10}(x) = 3$

To eliminate the log based 10 on the left, we will need to raise both terms as exponents of base 10.

$10^{log_{10}(x)} = 10^3$

x=1000

The answer is: 1000

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Example Question #41 : Solving Logarithms

Find if $log_{3}(x)=4$ .

Possible Answers:

Correct answer:

Explanation:

First, we start with the full equation:

$log_{3}(x)=4$

Now we can expand the right side of the equation:

$log_{3}(x)=1+1+1+1$

A log of it's own base equals :

$log_{3}(x)=log_{3}(3)+log_{3}(3)+log_{3}(3)+log_{3}(3)$

Now we add the logs on the right side of the equation by multiplying the terms inside the logs:

$log_{3}(x)=log_{3}(3*3*3*3)$

$log_{3}(x)=log_{3}(81)$

x=81

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Example Question #181 : Logarithms

Solve $log_{7}(x-4)=0$

Possible Answers:

Correct answer:

Explanation:

By definition, a logarithm of any base that has the term inside is equal to . So we set that term equal to :

x-4=1

x=5

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Example Question #41 : Solving And Graphing Logarithms

Solve $log_{5}(2x)=1$ .

Possible Answers:

$\frac{2}{5}$

$\frac{5}{2}$

Correct answer:

$\frac{5}{2}$

Explanation:

When a logarithm equals , the equation in the logarithm equals the logarithms base:

2x=5

$x=\frac{5}{2}$

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Algebra II : Algebra II

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #31 : Solving And Graphing Logarithms

Example Question #32 : Solving Logarithms

Example Question #33 : Solving Logarithms

Example Question #34 : Solving Logarithms

Example Question #38 : Solving Logarithms

Example Question #35 : Solving Logarithms

Example Question #36 : Solving Logarithms

Example Question #41 : Solving Logarithms

Example Question #181 : Logarithms

Example Question #41 : Solving And Graphing Logarithms

All Algebra II Resources

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