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Algebra II : Algebra II

Study concepts, example questions & explanations for Algebra II

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Example Questions

Example Question #3061 : Algebra Ii

Solve the equation:

$5e^{3x+4}-10=0$

Possible Answers:

x=-4+ln2

$x=\frac{ln6}{3}$

No solution exists

$x=\frac{-4+ln2}{3}$

$x=-\frac{4}{3}+ln10$

Correct answer:

$x=\frac{-4+ln2}{3}$

Explanation:

Get all the terms with e on one side of the equation and constants on the other.

$5e^{3x+4}-10=0$

$e^{3x+4}=2$

Apply the logarithmic function to both sides of the equation.

$ln(e^{3x+4})=ln2$

3x+4=ln2

3x=-4+ln2

$x=\frac{-4+ln2}{3}$

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Example Question #3062 : Algebra Ii

Solve the equation:

2log(x)+log(5)=3

Possible Answers:

$x=\pm \sqrt{995}$

$x=\pm6\sqrt{2}$

$x=\pm 12$

$x=\pm3\sqrt{5}$

$x=\pm 10\sqrt{2}$

Correct answer:

$x=\pm 10\sqrt{2}$

Explanation:

Recall the rules of logs to solve this problem.

First, when there is a coefficient in front of log, this is the same as log with the inside term raised to the outside coefficient.

2log(x)+log(5)=3

$log(x^{2})+log(5)=3$

Also, when logs of the same base are added together, that is the same as the two inside terms multiplied together.

In mathematical terms:

$log(a)+log(b)=log(a\cdot b)$

Thus our equation becomes,

$log(5x^{2})=3$

To simplify further use the rule,

$log_b(x)\rightarrow b^{log_b(x)}=x$

$10^{log(5x^{2})}=10^{3}$

$5x^{2}=1000$

$x^{2}=200$

$x=\pm10\sqrt{2}$ .

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Example Question #16 : Solving Logarithms

Solve for $\small x$ :

$\small \log_{2}(x+6)=4$

Possible Answers:

$\textup{No solution}$

$\frac{1}{4}$

Correct answer:

Explanation:

Logarithms are another way of writing exponents. In the general case, $\small \log_a{x}=b$ really just means $\small a^{b}=x$ . We take the base of the logarithm (in our case, 2), raise it to whatever is on the other side of the equal sign (in our case, 4) and set that equal to what is inside the parentheses of the logarithm (in our case, x+6). Translating, we convert our original logarithm equation into $\small 2^{4}=x+6$ . The left side of the equation yields 16, thus $\small x=10$ .

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Example Question #3063 : Algebra Ii

$\log_{5}x=3$

Possible Answers:

$\frac{1}{125}$

$\frac{1}{25}$

125

Correct answer:

125

Explanation:

To solve this equation, remember log rules

$\log_{b}x=y; b^{y}=x$ .

This rule can be applied here so that

5^3=x

and

125=x.

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Example Question #11 : Solving And Graphing Logarithms

$\log_{4}64-\log_{3}81=$

Possible Answers:

Correct answer:

Explanation:

To solve this equation, you must first simplify the log expressions and remember log laws ( $\log_{b}x=y$ ; $b^{y}=x$ ). Therefore, for the first expression, $4^{x}=64$ so x=3 . For the second expression, 3^x=81 so x=4. Then, substitute those numbers in so that you get 3-4=-1 .

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Example Question #19 : Solving Logarithms

Solve for

$log_{5}(x) = 2$

Possible Answers:

x = 12

x = 25

x = 32

x = 15

x = 5

Correct answer:

x = 25

Explanation:

$log_{5}(x) = 2$

Remember that a logarithm is nothing more than an exponent. The equation reads, the logarithm (exponent) with base that gives is .

x = 5^2

x = 25

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Example Question #20 : Solving Logarithms

Solve the equation

$\log_4(4x+7)=\log_4(11x)$

Possible Answers:

None of the other answers

x=2

x=3

x=0.5

x=1

Correct answer:

x=1

Explanation:

Since both sides are a log with the same base, their arguments must be equal.

Setting them equal:

4x+7=11x

Solve for x:

7=7x

1=x

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Example Question #21 : Solving Logarithms

Solve the equation

$\ln(3x-2)=\ln(6x)$

Possible Answers:

x=1

No Solution

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

x=2

$x=\frac{-2}{3}$

Correct answer:

No Solution

Explanation:

Set the arguments of the natural log equal and solve for

3x-2=6x

-2=3x

$\frac{-2}{3}=x$

However, you cannot take the natural log of a negative number. Therefore; there is No Solution

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Example Question #22 : Solving Logarithms

Solve:

$\log_{2}24$

Possible Answers:

$\log24$

$\log12$

$\frac{\log24}{\log2}$

$-\frac{\log24}{\log2}$

Correct answer:

$\frac{\log24}{\log2}$

Explanation:

To solve this problem, it is easiest to start by thinking about what the logarithm means:

2^x=24

Now, we take the log of both sides (either base-10 or natural):

$\log(2^x)=\log24$

Next, using the properties of logarithms, we can bring the exponent x in front of the logarithm:

$x\log2=\log24$

Finally, solve for x:

$x=\frac{\log24}{\log2}$

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Example Question #23 : Solving Logarithms

Evaluate

$\log_5 25 = x$

Possible Answers:

Correct answer:

Explanation:

Logarithms are inverses of exponents.

$\log_5 25$ is asking how many fives multiplied together are equal to 25.

The formula to solve logarithmic equations is as follows.

$log_ab=c\Rightarrow a^c=b$

Applying this formula to our particular problem results in the following.

$\log_5 25\Rightarrow 5^x=25$

5^2 = 5*5 = 25

So x= 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 #3061 : Algebra Ii

Example Question #3062 : Algebra Ii

Example Question #16 : Solving Logarithms

Example Question #3063 : Algebra Ii

Example Question #11 : Solving And Graphing Logarithms

Example Question #19 : Solving Logarithms

Example Question #20 : Solving Logarithms

Example Question #21 : Solving Logarithms

Example Question #22 : Solving Logarithms

Example Question #23 : Solving Logarithms

All Algebra II Resources

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