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Algebra II : Solving and Graphing Logarithms

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Example Questions

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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Example Question #3101 : Algebra Ii

Solve log_x(16)=4

Possible Answers:

Correct answer:

Explanation:

First we rewrite the equation in exponential form:

x^4=16

Now we take the cube root of :

x=2

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Example Question #54 : Solving Logarithms

Solve 3+3log(2x)=12 .

Possible Answers:

5000

500

Correct answer:

500

Explanation:

First we subtract from both sides:

3log(2x)=9

Then we divide both sides by :

log(2x)=3

Now it would help if we wrote the equation in exponential form (remember, if the log doesn't show a base, it's base 10):

2x=10^3

Finally, we use algebra to solve:

2x=1000

x=500

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Example Question #201 : Logarithms

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

Possible Answers:

Correct answer:

Explanation:

First, we can combine the log terms:

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

Now we can change to exponent form (remember, if a log doesn't specifically have a base, then it's base 10):

(x+2)(x-1)=10^1

We need to set the equation equal to in order to solve the quadratic equation, so we combine the terms and subtract :

x^2+x-12=0

Then we factor and solve for :

(x+4)(x-3)=0

x=-4, 3

Lastly, we have to check our answers. When we plug in to the original equation, everything comes out well. However, when we use we get errors (because you can't take the log of a negative number). Therefore, we only have 1 solution: x=3 .

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Example Question #202 : Logarithms

Solve log(x)+log(x-48)=2 .

Possible Answers:

Correct answer:

Explanation:

First, we combine the logs:

log((x)(x-48))=2

Next, we can rewrite the equation in exponent form:

x(x-48)=10^2

Now we use algebra to set the equation equal to :

x^2-48x-100=0

We then factor and solve for :

(x-50)(x+2)=0

x=50, -2

Lastly, we have to check our answers. Using in the original equation results in errors (because you can't take the log of a negative number). works out fine though, so we have 1 solution.

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Example Question #52 : Solving Logarithms

Solve -9+6log(8x)=27

Possible Answers:

124

125000

Correct answer:

125000

Explanation:

First, we add to each side of the equation:

6log(8x)=36

Next, we divide each side by :

log(8x)=6

Now, we write the equation in exponent form and solve for :

8x=10^6

$x=\frac{10^6}{8}$

x=125000

Plugging our answer doesn't produce any errors, so it's a valid answer!

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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 #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

Example Question #3101 : Algebra Ii

Example Question #54 : Solving Logarithms

Example Question #201 : Logarithms

Example Question #202 : Logarithms

Example Question #52 : Solving Logarithms

All Algebra II Resources

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