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Algebra II : Logarithms

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Example Questions

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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Example Question #161 : Logarithms

Solve:

$\log_84+\log_810$

Possible Answers:

$\frac{\log40}{\log8}$

$\log5$

$\frac{\log14}{\log8}$

$\log32$

Correct answer:

$\frac{\log40}{\log8}$

Explanation:

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

$\log_ab$ means 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):

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

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

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:

$\log8^x=\log40$

$x\log8=\log40$

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

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Example Question #25 : Solving Logarithms

Solve for x:

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

Possible Answers:

$\log\frac{3}{4}$

$\frac{\log4}{\log3}$

$\frac{\log3}{\log4}$

$\log1$

$\log\frac{4}{3}$

Correct answer:

$\frac{\log4}{\log3}$

Explanation:

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

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

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

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

Now, solve for x:

$\log4=x\log3$

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

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Example Question #161 : Logarithms

If $log_{m}(2)=.387$ and $log_{m}(3)=.613$ , what is ?

Possible Answers:

Correct answer:

Explanation:

First, we know from logarithmic properties that:

$log_{n}(n)=1$

So if we can find a combination of our knowns that equal , we should be able to figure out . It may take some guess and check, but:

.387 + .613 = 1

That would equate to:

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

Using our log properties, we can simplify into:

$log_{m}(2\cdot 3)=1$

$log_{m}(6)=1$

That means that:

m=6

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Example Question #27 : Solving Logarithms

Evaluate $log_{100}(10000)$ .

Possible Answers:

100

Correct answer:

Explanation:

There are a few ways to go about this, but let's use a change of base to make the problem easier to work without using a calculator. First, we know that:

$log_{m}(n)=\frac{log_x(m)}{log_x(n)}$

We can choose any value for that we would like here, so to make things simple, let's put this problem in base :

$log_{100}(10000) = \frac{log_{10}(10000)}{log_{10}(100)} = \frac{4}{2} = 2$

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Algebra II : Logarithms

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

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

Example Question #161 : Logarithms

Example Question #25 : Solving Logarithms

Example Question #161 : Logarithms

Example Question #27 : Solving Logarithms

All Algebra II Resources

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