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Algebra II : Mathematical Relationships and Basic Graphs

Study concepts, example questions & explanations for Algebra II

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Example Questions

Example Question #9 : Natural Log

Determine the value of: $2e\cdot ln(e^3)$

Possible Answers:

6e^5

e^6

e^7

Correct answer:

Explanation:

In order to simplify this expression, use the following natural log rule.

ln_x(x^y) =y

The natural log has a default base of . This means that:

$2e\cdot ln(e^3)=2e\cdot ln_e(e^3) = 2e\cdot (3)$

The answer is:

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Example Question #4 : Natural Log

Simplify: $2e^{4ln{(2e)}}$

Possible Answers:

32e^4

32e^2

64e^4

16e

16e^4

Correct answer:

32e^4

Explanation:

According to log properties, the coefficient in front of the natural log can be rewritten as the exponent raised by the quantity inside the log.

$2e^{4ln{(2e)}} = 2e^{ln[{(2e)^4}]}$

Notice that natural log has a base of . This means that raising the log by base will eliminate both the and the natural log.

The terms become: 2(2e)^4

Simplify the power.

2(2e)^4 = 2(16e^4) = 32e^4

The answer is: 32e^4

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Example Question #11 : Natural Log

Evaluate: 2e^2ln(e^2)

Possible Answers:

4e^4

4e^2

Correct answer:

4e^2

Explanation:

The natural log has a default base of . Natural log to of an exponential raised to the power will be just the power. The natural log and will be eliminated.

ln(e^x)=x

Rewrite the expression.

2e^2ln(e^2) = 2e^2(2) = 4e^2

The answer is: 4e^2

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Example Question #11 : Natural Log

Determine the value of: $6ln(e^{-3x})$

Possible Answers:

-18x

-3x+6

-216x

$-\frac{1}{2}x$

-9x

Correct answer:

-18x

Explanation:

The natural log has a default base of .

According to the rule of logs, we can use:

$log_{b}(b^x) = x$

$6ln(e^{-3x}) = 6ln_e(e^{-3x})$

The coefficient in front of the natural log can be transferred as the power of the exponent.

$6ln_e(e^{-3x}) = ln_e(e^{(-3x)(6)})$

The natural log and base e will cancel, leaving just the exponent.

(-3x)(6)=-18x

The answer is: -18x

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Example Question #11 : Understanding Logarithms

Solve: $3ln(e^{7e})$

Possible Answers:

21e

7e^2

21e^2

7e^3

Correct answer:

21e

Explanation:

The natural log has a default base of . This means that the natural log of to the certain power will be just the power itself.

$log_{b}(b^x) = x$

The expression $3ln(e^{7e})$ becomes:

3(7e) = 21e

The answer is: 21e

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Example Question #14 : Natural Log

Simplify: ln(6e^7)

Possible Answers:

ln(42e)

ln(2)+ln(3)+7

42e

ln(5)+e^7

Correct answer:

ln(2)+ln(3)+7

Explanation:

Use the log properties to separate each term. When the terms inside are multiplied, the logs can be added.

Rewrite the expression.

ln(6e^7) = ln(6)+ln(e^7) = ln(2)+ln(3)+ln(e^7)

The exponent, 7 can be dropped as the coefficient in front of the natural log. Natural log of the exponential is equal to one since the natural log has a default base of .

$log_{b}(b^x) = x$

The answer is: ln(2)+ln(3)+7

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Example Question #15 : Natural Log

The equation $kt = ln(\frac{T-S}{T_0-S})$ represents Newton's Law of Cooling. Solve for T_0 .

Possible Answers:

$T_0 =\frac{e^{kt}S-T}{e^{kt}+S}$

$T_0 =\frac{T-S+e^{kt}S}{e^{kt}}$

$T_0 =\frac{T+e^{kt}S}{e^{kt}+S}$

$T_0 =\frac{T-e^{kt}S}{e^{kt}+S}$

$T_0 =\frac{e^{kt}S}{e^{kt}+S+T}$

Correct answer:

$T_0 =\frac{T-S+e^{kt}S}{e^{kt}}$

Explanation:

Use base and raise both the left and right sides as the powers of . This will eliminate the natural log term.

$e^{kt} =e^ {ln(\frac{T-S}{T_0-S})}$

The equation becomes:

$e^{kt} =\frac{T-S}{T_0-S}$

Multiply the quantity (T_0-S) on both sides.

$e^{kt}(T_0-S) =\frac{T-S}{T_0-S}(T_0-S)$

$e^{kt}T_0-e^{kt}S =T-S$

Add $e^{kt}S$ on both sides.

$e^{kt}T_0-e^{kt}S +e^{kt}S=T-S+e^{kt}S$

The equation becomes:

$e^{kt}T_0=T-S+e^{kt}S$

To isolate T_0 , divide $e^{kt}$ on both sides.

$\frac{e^{kt}T_0}{e^{kt}}=\frac{T-S+e^{kt}S}{e^{kt}}$

The answer is: $T_0 =\frac{T-S+e^{kt}S}{e^{kt}}$

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Example Question #16 : Natural Log

Evaluate: $2e^2ln(e^{2e})$

Possible Answers:

2e^2+2e

e+4

4e^2+e

4e^3

Correct answer:

4e^3

Explanation:

The natural log has a default base of .

Use the log property:

log_x(x^y)=y

We can cancel the base and the log of the base.

The expression $2e^2ln(e^{2e})$ becomes:

2e^2(2e )= 4e^3

The answer is: 4e^3

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Example Question #11 : Understanding Logarithms

Simplify: ln(e+e^2)

Possible Answers:

1+ln(e+1)

2e+1

Correct answer:

1+ln(e+1)

Explanation:

Notice that the terms inside the log are added, with a common factor of . Pull out a common factor.

ln(e+e^2)= ln((e)(1+e))

Notice that these two terms inside the log are multiplied. We can split the log into two terms.

ln((e)(1+e)) = ln(e)+ln(e+1)

The value of the first term is equal to one, since natural log has a default base of . We can use the property ln_xx^y=y to eliminate the log and the term, which will cancel leaving just the power of one.

The expression becomes: 1+ln(e+1)

We cannot use the property of logs to simplify the second term.

The answer is: 1+ln(e+1)

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Example Question #18 : Natural Log

Solve the expression: ln(2x+e)= 13

Possible Answers:

$\frac{e^{13}-e}{2}$

$\frac{e^{13}+e}{2}$

$\frac{e^{12}}{2}$

$\frac{e^6}{2}$

$\frac{13-e}{2e}$

Correct answer:

$\frac{e^{13}-e}{2}$

Explanation:

In order to eliminate the natural log and solve for x, we will need to exponential both sides because is the base of natural log.

$e^{ln(2x+e)}= e^{13}$

The left side will be reduced to just the inner quantity of the natural log.

$2x+e = e^{13}$

Subtract from both sides of the equation.

$2x+e -e= e^{13}-e$

$2x= e^{13}-e$

Divide by two on both sides.

$\frac{2x}{2}= \frac{e^{13}-e}{2}$

The answer is: $\frac{e^{13}-e}{2}$

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Algebra II : Mathematical Relationships and Basic Graphs

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #9 : Natural Log

Example Question #4 : Natural Log

Example Question #11 : Natural Log

Example Question #11 : Natural Log

Example Question #11 : Understanding Logarithms

Example Question #14 : Natural Log

Example Question #15 : Natural Log

Example Question #16 : Natural Log

Example Question #11 : Understanding Logarithms

Example Question #18 : Natural Log

All Algebra II Resources

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