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Algebra II : Mathematical Relationships and Basic Graphs

Study concepts, example questions & explanations for Algebra II

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Example Questions

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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Example Question #3097 : Algebra Ii

Solve log (3 x + 1) - 2 log (x) = 1 .

Possible Answers:

$-\frac{1}{5}$

$\frac{1}{4}$

$\frac{1}{e}$

$\frac{1}{2}$

$\frac{1}{10}$

Correct answer:

$\frac{1}{2}$

Explanation:

The first thing we can do is write the coefficients in front of the logs as exponents of the terms inside the logs:

log (3 x + 1) - log (x^2) = 1

Next, we combine the logs:

$log(\frac{3x+1}{x^2})=1$

Now we can change the equation into exponent form:

$\frac{(3x+1)}{x^2}=10^1$

When we simplify, we'll also move all the terms to one side of the equation:

10x^2-3x-1=0

From here, we factor to get our solutions:

(5x+1)(2x-1)=0

$x=\frac{1}{2}, -\frac{1}{5}$

The last thing we have to do is check our answers. $\frac{1}{2}$ doesn't raise any problems, but $-\frac{1}{5}$ does. If we plug it back into the original equation, we would be evaluating a negative value in a log, which we can't do.

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Example Question #3098 : Algebra Ii

Solve: $2log_{2}(64)-3log_{4}(64)$

Possible Answers:

Correct answer:

Explanation:

In order to solve for the logs, we will need to write the log properties as follows:

x^y = z and $log_{x}z = y$

This means that:

$log_{2}(64) = 6$

$log_{4}(64) = 3$

Replace the values into the expression.

$2log_{2}(64)-3log_{4}(64)= 2(6)-3(3) = 12-9 =3$

The answer is:

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Example Question #3099 : Algebra Ii

Solve $ln (e^{5x}) = 2$

Possible Answers:

$\frac{2}{5}$

$\frac{5}{2}$

$\frac{1}{e}$

Correct answer:

$\frac{2}{5}$

Explanation:

First, we rewrite the equation as:

$log_e(e^{5x})=2$

We can use log properties to simplify:

5x=2

From here, simple algebra is used to solve:

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

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Example Question #431 : Mathematical Relationships And Basic Graphs

Solve 2 log_x(64)=3 .

Possible Answers:

Correct answer:

Explanation:

First we divide both sides of the equation by :

$log_x(64)=\frac{3}{2}$

Next, we write the equation in exponential form:

$x^{\frac{3}{2}}=64$

Then we solve for by taking the cube root, and then squaring :

x=16

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Example Question #52 : Solving Logarithms

Solve $2 log_{(x+1)}(9)=4$ .

Possible Answers:

Correct answer:

Explanation:

First we can divide each side by :

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

Next, we rewrite in exponent form:

(x+1)^2=9

From here, we simply take the square root of , and then subtract :

x=2

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Algebra II : Mathematical Relationships and Basic Graphs

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

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

Example Question #3097 : Algebra Ii

Example Question #3098 : Algebra Ii

Example Question #3099 : Algebra Ii

Example Question #431 : Mathematical Relationships And Basic Graphs

Example Question #52 : Solving Logarithms

All Algebra II Resources

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