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Algebra II : Algebra II

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Example Questions

Example Question #2961 : Algebra Ii

Convert the following logarithmic equation to an exponential equation:

$log_{13}b=147$

Possible Answers:

147^b=13

$13^{147}=b$

$b^{13}=147$

$13^{b}=147$

Correct answer:

$13^{147}=b$

Explanation:

Convert the following logarithmic equation to an exponential equation:

$log_{13}b=147$

Recall the following:

This

log_b(a)=c

Can be rewritten as

b^c=a

So, our given logarithm

$log_{13}b=147$

Can be rewritten as

$13^{147}=b$

Fortunately we don't need to expand, because this woud be a very large number!

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Example Question #11 : Logarithms And Exponents

Rewrite the follwing equation as a logarithm:

14^b=977

Possible Answers:

$log_{b}977=14$

$log_{977}14=b$

$log_{14}b=977$

$log_{14}977=b$

Correct answer:

$log_{14}977=b$

Explanation:

Rewrite the follwing equation as a logarithm:

14^b=977

To complete this problem, recall the following relationship:

can be rewritten as b^c=a

So, this:

14^b=977

Is the same thing as this:

$log_{14}977=b$

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Example Question #2961 : Algebra Ii

Simplify: 2ln(e^3)

Possible Answers:

e^6

e^5

2e^6

Correct answer:

Explanation:

When the base is raised to a certain power, taking the natural log of this whole term will eliminate the exponential and the power can be pulled out as the coefficient.

2ln(e^3) = 2(3) ln(e^1) = 2(3)= 6

The answer is:

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Example Question #12 : Logarithms And Exponents

Solve: $log_{\frac{1}{5}} 125$

Possible Answers:

$-\frac{3}{5}$

$-\frac{1}{3}$

$-\frac{3}{125}$

$-\frac{5}{3}$

Correct answer:

Explanation:

In order to solve this log, we will need to write 125 in terms of one fifth to a certain power.

Rewrite 125 as an exponent of one-fifth.

$125 = (\frac{1}{5})^{-3}$

$log_{\frac{1}{5}} 125=log_{\frac{1}{5}}(\frac{1}{5})^{-3}$

According to the log rule, $log_{x}x^y=y$ , the bases will cancel, leaving just the exponent.

The answer is:

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Example Question #2964 : Algebra Ii

Simplify, if possible: 2log(125)

Possible Answers:

5log(5)

6log(3)

10log(3)

6log(5)

$\textup{The answer is already simplified.}$

Correct answer:

6log(5)

Explanation:

Notice that the term in the log can be rewritten as a base raised to a certain power.

Rewrite the number in terms of base five.

2log(125)=2log(5^3)

According to log rules, the exponent can be dropped as a coefficient in front of the log.

$2\cdot 3log(5) = 6log(5)$

The answer is: 6log(5)

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Example Question #2965 : Algebra Ii

Solve: $3log_{5}(5^{10})$

Possible Answers:

1000

300

Correct answer:

Explanation:

Evaluate the log using the following property:

log_xx^y = y

The log based and the base of the term will simplify.

The expression becomes:

$3log_{5}(5^{10}) = 3(10) = 30$

The answer is:

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Example Question #2962 : Algebra Ii

Try to answer without a calculator.

True or false:

$\log_{7} 700 = 2$

Possible Answers:

True

False

Correct answer:

False

Explanation:

By definition, $\log_{7} 700 = 2$ if and only if $7^{2} = 700$ . However,

$7^{2} = 7 \times 7 = 49 \ne 700$ ,

making this false.

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Example Question #64 : Understanding Logarithms

Try without a calculator:

Evaluate $\log_{16} 8$

Possible Answers:

$-\frac{3}{4}$

$\frac{4}{3}$

None of the other choices gives the correct response.

$\frac{3}{4}$

$-\frac{4}{3}$

Correct answer:

$\frac{3}{4}$

Explanation:

By definition, $\log_{16} 8 = N$ if and only if $16^{N} = 8$ .

8 and 16 are both powers of 2; specifically, $8= 2^{3} , 16 = 2^{4}$ . The latter equation can be rewritten as

$\left (2 ^{4} \right ) ^{N} = 2^{3}$

By the Power of a Power Property, the equation becomes

$2 ^{4\cdot N} = 2^{3}$

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

It follows that

4N = 3 ,

and

$N = \frac{3}{4}$ ,

the correct response.

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Example Question #2967 : Algebra Ii

Solve for (nearest tenth):

$\log_{7} N = 1.8$ .

Possible Answers:

2.9

12.6

61.2

33.2

Correct answer:

33.2

Explanation:

By definition, $\log_{a}N = M$ if and only if $a^{M}= N$ . Set a= 7, M = 1.8 , and

$\log_{7} N = 1.8$

if and only if

$N= 7^{1.8}$

Through calculation, we see that

$N \approx 33.2$ .

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Example Question #2963 : Algebra Ii

Try to answer without a calculator:

Which is true about $\log_{0}1$ ?

Possible Answers:

Cannot be determined

$\log_{0}1$ is an undefined quantity

$\log_{0}1 = 1$

$\log_{0}1 = 0$

Correct answer:

$\log_{0}1$ is an undefined quantity

Explanation:

The question asks for the value of the "base 0 logarithm" of 0. However, this is not defined, as a logarithm can only have as its base a positive number not equal to 1.

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Algebra II : Algebra II

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #2961 : Algebra Ii

Example Question #11 : Logarithms And Exponents

Example Question #2961 : Algebra Ii

Example Question #12 : Logarithms And Exponents

Example Question #2964 : Algebra Ii

Example Question #2965 : Algebra Ii

Example Question #2962 : Algebra Ii

Example Question #64 : Understanding Logarithms

Example Question #2967 : Algebra Ii

Example Question #2963 : Algebra Ii

All Algebra II Resources

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