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Algebra II : Solving and Graphing Logarithms

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Example Questions

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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Example Question #42 : Solving And Graphing Logarithms

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

Possible Answers:

Correct answer:

Explanation:

Rearranging the logarithm so that we exchange an exponent for the log we get:

$10^{-1}=x$

x=.1

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Example Question #43 : Solving And Graphing Logarithms

Solve $log_{8}(2x+4)=1$ .

Possible Answers:

$-\frac{3}{2}$

Correct answer:

Explanation:

First we rearrange the equation, trading the logarithm for an exponent:

8^1=2x+4

And then we solve:

4=2x

2=x

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Example Question #3093 : Algebra Ii

Solve $log_{3}(x+1)+log_{3}(x)=2$ .

Possible Answers:

Correct answer:

Explanation:

The first thing we can do is combine the log terms:

$log_{3}((x+1)x)=2$

Now we can change to exponent form:

3^2=(x+1)x

We can combine terms and set the equation equal to to have a quadratic equation:

x^2+x-6=0

We then solve the equation and get the answers:

and

can't be an answer, because the values inside a log can't be negative, so that leaves us with a single answer of .

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Example Question #3093 : Algebra Ii

Solve log_2 (x + 1) +log_2 (x - 1) = 3 .

Possible Answers:

Correct answer:

Explanation:

The first thing we can do is combine log terms:

log_2((x+1)(x-1))=3

Simplifying the log term gives:

log_2(x^2-1)=3

Now we can change the equation to exponent form:

x^2 - 1 = 2^3

And to solve:

x^2-1=8

x^2=9

x=3, x-3

Here, the solution can't be because the term inside a logarithm can't be negative, so the only solution is .

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Example Question #3094 : Algebra Ii

Solve log_2 (x^2 + 7) = 5

Possible Answers:

3, -7

5, -5

4, -6

Correct answer:

5, -5

Explanation:

First, let's change the equation to exponent form:

x^2+7=2^5

Then simplify:

x^2+7=32

And solve:

x^2=25

x=5, x=-5

Both answers are valid because in the original equation is squared, so any negative numbers don't cause the logarithm to become negative.

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Example Question #3095 : Algebra Ii

Solve log_6(4x)=log_6(x+9)

Possible Answers:

Correct answer:

Explanation:

We can start by getting both the log terms on the same side of the equation:

log_6(4x)-log_6(x+9)=0

Then we combine log terms:

$log_6(\frac{4x}{x+9})=0\,$

Now we can change to exponent form:

$\frac{4x}{x+9}=6^0$

Anything raised to the th power equals , and from here it becomes a simpler problem to solve:

$\frac{4x}{x+9}=1$

4x=x+9

3x=9

x=3

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Example Question #3096 : Algebra Ii

Solve ln(x-2) + ln(x) = ln(2x+9)

Possible Answers:

Correct answer:

Explanation:

First we're going to get all the natural logs on one side of the equation:

ln(x-2) + ln(x) - ln(2x+9)= 0

Next, we're going to combine all the terms into one natural log:

$ln(\frac{(x-2)(x)}{2x+9}) = 0$

Now we can change to exponent form:

$\frac{(x-2)(x)}{2x+9} = e^0$

Anything raised to the th power equals , which helps us simplify:

$\frac{(x-2)(x)}{2x+9} = 1$

(x-2)(x) = 2x+9

x^2-2x = 2x+9

x^2-9=0

From here, we can factor and solve:

(x+3)(x-3)=0

x=3, -3

We have to notice, however, that isn't a valid answer because if we were to plug it into the original formula we would have a negative value in a logarithm.

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Algebra II : Solving and Graphing Logarithms

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #41 : Solving Logarithms

Example Question #181 : Logarithms

Example Question #41 : Solving And Graphing Logarithms

Example Question #42 : Solving And Graphing Logarithms

Example Question #43 : Solving And Graphing Logarithms

Example Question #3093 : Algebra Ii

Example Question #3093 : Algebra Ii

Example Question #3094 : Algebra Ii

Example Question #3095 : Algebra Ii

Example Question #3096 : Algebra Ii

All Algebra II Resources

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