Algebra II : Mathematical Relationships and Basic Graphs

Study concepts, example questions & explanations for Algebra II

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Example Questions

Example Question #154 : Logarithms

Solve for \displaystyle \small x

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

Possible Answers:

\displaystyle \frac{1}{4}

\displaystyle \textup{No solution}

\displaystyle 2

\displaystyle 10

\displaystyle 0

Correct answer:

\displaystyle 10

Explanation:

Logarithms are another way of writing exponents. In the general case, \displaystyle \small \log_a{x}=b really just means \displaystyle \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 \displaystyle \small 2^{4}=x+6. The left side of the equation yields 16, thus \displaystyle \small x=10.

Example Question #61 : Exponential And Logarithmic Functions

\displaystyle \log_{5}x=3

Possible Answers:

\displaystyle \frac{1}{125}

\displaystyle 1

\displaystyle 125

\displaystyle 25

\displaystyle \frac{1}{25}

Correct answer:

\displaystyle 125

Explanation:

To solve this equation, remember log rules

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

This rule can be applied here so that

\displaystyle 5^3=x

and

\displaystyle 125=x.

Example Question #161 : Logarithms

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

Possible Answers:

\displaystyle 2

\displaystyle 3

\displaystyle 1

\displaystyle -1

\displaystyle -4

Correct answer:

\displaystyle -1

Explanation:

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

Example Question #162 : Logarithms

Solve for \displaystyle x

\displaystyle log_{5}(x) = 2

Possible Answers:

\displaystyle x = 12

\displaystyle x = 32

\displaystyle x = 5

\displaystyle x = 25

\displaystyle x = 15

Correct answer:

\displaystyle x = 25

Explanation:

\displaystyle log_{5}(x) = 2

 

Remember that a logarithm is nothing more than an exponent. The equation reads, the logarithm (exponent) with base \displaystyle 5 that gives \displaystyle x is \displaystyle 2

 

\displaystyle x = 5^2

 

\displaystyle x = 25

Example Question #163 : Logarithms

Solve the equation

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

Possible Answers:

None of the other answers

\displaystyle x=2

\displaystyle x=3

\displaystyle x=0.5

\displaystyle x=1

Correct answer:

\displaystyle x=1

Explanation:

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

Setting them equal:

\displaystyle 4x+7=11x

Solve for x:

\displaystyle 7=7x

\displaystyle 1=x

Example Question #164 : Logarithms

Solve the equation

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

Possible Answers:

No Solution

\displaystyle x=2

\displaystyle x=1

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

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

Correct answer:

No Solution

Explanation:

Set the arguments of the natural log equal and solve for \displaystyle x

\displaystyle 3x-2=6x

\displaystyle -2=3x

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

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

Example Question #165 : Logarithms

Solve:

\displaystyle \log_{2}24

Possible Answers:

\displaystyle \frac{\log24}{\log2}

\displaystyle -\frac{\log24}{\log2}

\displaystyle \log24

\displaystyle \log12

Correct answer:

\displaystyle \frac{\log24}{\log2}

Explanation:

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

\displaystyle 2^x=24

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

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

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

\displaystyle x\log2=\log24

Finally, solve for x:

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

Example Question #163 : Logarithms

Evaluate

\displaystyle \log_5 25 = x

Possible Answers:

\displaystyle 2

\displaystyle 3

\displaystyle 1

\displaystyle 5

\displaystyle 4

Correct answer:

\displaystyle 2

Explanation:

Logarithms are inverses of exponents. 

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

The formula to solve logarithmic equations is as follows.

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

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

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

\displaystyle 5^2 = \displaystyle 5*5 = 25

So \displaystyle x= 2.

Example Question #21 : Solving Logarithms

Solve:

\displaystyle \log_84+\log_810

Possible Answers:

\displaystyle \log5

\displaystyle \frac{\log40}{\log8}

\displaystyle \frac{\log14}{\log8}

\displaystyle \log32

Correct answer:

\displaystyle \frac{\log40}{\log8}

Explanation:

To solve, we must keep in mind that a logarithm is asking the following:

\displaystyle \log_ab means \displaystyle a^x=b, and we must find x.

First, we can rewrite the sum of the base-8 logarithms as a product (this is a property of logarithms):

\displaystyle \log_8(4\cdot 10)=\log_840

Now, we use the above formula to find what this numerically equals:

\displaystyle 8^x=40

Take the natural or common logarithm of both sides, which allows us to bring the power x in front of the logarithm:

\displaystyle \log8^x=\log40

\displaystyle x\log8=\log40

\displaystyle x=\frac{\log40}{\log8}

Example Question #167 : Logarithms

Solve for x:

\displaystyle 4^x=3^{x^2}

Possible Answers:

\displaystyle \log\frac{4}{3}

\displaystyle \log\frac{3}{4}

\displaystyle \frac{\log4}{\log3}

\displaystyle \frac{\log3}{\log4}

\displaystyle \log1

Correct answer:

\displaystyle \frac{\log4}{\log3}

Explanation:

To solve for x, we must take the logarithm of both sides (common or natural, it doesn't matter):

\displaystyle \log(4^x)=\log(3^{x^2})

In doing this, we can now bring the exponents in front of the logarithms:

\displaystyle x\log4=x^2\cdot\log3

Now, solve for x:

\displaystyle \log4=x\log3

\displaystyle x=\frac{\log4}{\log3}

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