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Algebra II : Logarithms

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Example Questions

Example Question #28 : Solving Logarithms

Evaluate log_3(27) .

Possible Answers:

Correct answer:

Explanation:

We can take this problem and expand it a bit, which will make things easier in the long run. We know that:

$log_3(27)=log_3(3\cdot 3\cdot 3)$

Using one of our logarithmic properties, we can expand even further:

$log_3(3\cdot 3\cdot 3)=log_3(3)+log_3(3)+log_3(3)$

Another log property states that:

log_x(x)=1

So:

log_3(3)+log_3(3)+log_3(3) = 1+1+1 = 3

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Example Question #21 : Solving And Graphing Logarithms

Solve:

$\log_4 16^x=10$

Possible Answers:

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

x=5

x=1

x=0

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

Correct answer:

x=5

Explanation:

To solve for x, remember that exponents inside logarithms can be moved to the front of the logarithm:

$x\cdot \log_416=10$

Next, we can rewrite the logarithm as the number it equals, and solve for x:

$x\cdot2=10$

x=5

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Example Question #22 : Solving And Graphing Logarithms

Solve: ln(2x) = 100

Possible Answers:

$e^{50}$

$\frac{e^{100}}{2}$

$2e^{100}$

Correct answer:

$\frac{e^{100}}{2}$

Explanation:

In order to solve for the x-variable, we will need to exponential function both sides of the equation in order to eliminate the natural log.

$e^{ln(2x) }= e^{100}$

The equation becomes:

$2x= e^{100}$

Divide by two on both sides.

$\frac{2x}{2}= \frac{e^{100}}{2}$

$x= \frac{e^{100}}{2}$

The answer is: $\frac{e^{100}}{2}$

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Example Question #3071 : Algebra Ii

Expand the log: $log_4(\frac{x^3}{y^5})$

Possible Answers:

$\frac{5}{3}log_4(\frac{x}{y})$

4log_3(x)-4log_5(y)

3log_4(x)-5log_4(y)

log_4(3x)-log_4(5y)

12log(x)-20log(y)

Correct answer:

3log_4(x)-5log_4(y)

Explanation:

Rewrite the logarithm using the quotient property.

$log_a(\frac{X}{Y}) = log_a(X)-log_a(Y)$

Rewrite the log.

$log_4(\frac{x^3}{y^5}) = log_4(x^3)-log_4(y^5)$

Use the power property to pull down the power.

log_a(X^n) = (n)log_a(X)

Simplify the log terms.

The answer is: 3log_4(x)-5log_4(y)

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Example Question #3072 : Algebra Ii

Solve for :

$log_{16}(-x)=4$

Possible Answers:

-16=x

None of these

x=144

-4=x

-65536=x

Correct answer:

-65536=x

Explanation:

$log_{16}(-x)=4$

Using definition of logarithms:

16^4=-x

65536=-x

-65536=x

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Example Question #3073 : Algebra Ii

Solve for :

$log_{16}(4)=-x$

Possible Answers:

x=-4

x=-.5

x=4

x=.5

None of these

Correct answer:

x=-.5

Explanation:

$log_{16}(4)=-x$

Using definition of logarithms

$16^{-x}=4$

$log 16^{-x}=log 4$

-x*log 16=log 4

$-x=\frac{log4}{log16}$

x=-.5

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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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Algebra II : Logarithms

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #28 : Solving Logarithms

Example Question #21 : Solving And Graphing Logarithms

Example Question #22 : Solving And Graphing Logarithms

Example Question #3071 : Algebra Ii

Example Question #3072 : Algebra Ii

Example Question #3073 : Algebra Ii

Example Question #31 : Solving And Graphing Logarithms

Example Question #32 : Solving Logarithms

Example Question #33 : Solving Logarithms

Example Question #34 : Solving Logarithms

All Algebra II Resources

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