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Algebra II : Logarithms

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Example Questions

Example Question #11 : Log Base 10

Simplify: $log(10^{-6}\cdot 100^{-3})$

Possible Answers:

$\textup{No solution.}$

Correct answer:

Explanation:

The log is in default base 10. To simplify this log, we will need to change the base of 100 to base 10.

100=10^2

Rewrite the inner quantity.

$log(10^{-6}\cdot 100^{-3})=log(10^{-6}\cdot 10^{2(-3)})=log(10^{-6}\cdot 10^{-6} )$

We can use the additive rule of exponents since both bases are the same.

$log(10^{-12}) = log_{10}(10^{-12})$

According to the rule of logs, a log of a base with similar bases will cancel, and will leave only the power.

$log_{x}x^y = y$

The answer is:

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Example Question #11 : Log Base 10

Evaluate: log(3x)= 3

Possible Answers:

$3e^{10}$

$\frac{ 1000}{3}$

3^9

$\frac{100}{3}$

$\frac{9}{1000}$

Correct answer:

$\frac{ 1000}{3}$

Explanation:

In order to eliminate the log, which has a default base of 10, we will need to raise both sides of the equation as powers using the value of 10.

$10^{log_{10}(3x)}= 10^3$

The equation becomes:

3x= 1000

Divide by three on both sides.

$\frac{3x}{3}=\frac{ 1000}{3}$

The answer is: $\frac{ 1000}{3}$

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Example Question #21 : Log Base 10

Solve: 2log(300)

Possible Answers:

4log(2)+2

2log(3)+4

3log(2)+4

2log(3)+2

3log(2)+2

Correct answer:

2log(3)+4

Explanation:

Break up log(300) using log rules. The log has a default base of ten.

$log(A\times B) = log(A)+log(B)$

$log(300) = log(3)+log(100)=log_{10}(3)+log_{10}(10^2)$

The exponent can be brought down as the coefficient since the bases of the second term are common.

$log_{10}(3)+log_{10}(10^2) =log_{10}(3)+2log_{10}(10) = log(3)+2$

This means that: 2log(300)=2[ log(3)+2] = 2log(3)+4

The answer is: 2log(3)+4

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Example Question #22 : Log Base 10

Evaluate: $2log_{10}(\frac{1}{1000})$

Possible Answers:

$\frac{2}{3}$

$\frac{1}{5}$

$\frac{1}{50}$

Correct answer:

Explanation:

We will need to write fraction in terms of the given base of log, which is ten.

$2log_{10}(\frac{1}{1000})=2log_{10}(10^{-3})$

According to the log rules:

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

This means that the expression of log based 10 and the power can be simplified.

$2log_{10}(10^{-3}) =2\cdot -3 = -6$

The answer is:

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Example Question #41 : Understanding Logarithms

Evaluate $\log_{8}172$ to the nearest tenth.

Possible Answers:

2.5

2.1

2.9

2.7

2.3

Correct answer:

2.5

Explanation:

Since most calculators only have common and natural logarithm keys, this can best be solved as follows:

By the Change of Base Property of Logarithms, if M,N,a > 0 and $a, M \ne 1$ ,

$\log_{M}N = \frac{\log_{a}N}{\log_{a}M}$

Setting M = 8, N = 172, a = 10 , we can restate this logarithm as the quotient of two common logarithms, and calculate accordingly:

$\log_{8}172 = \frac{\log 172}{\log 8}\approx \frac{2.2355 }{0.9031}\approx 2.4754$

or, when rounded, 2.5.

This can also be done with natural logarithms, yielding the same result.

$\log_{8}172 = \frac{\ln 172}{\ln 8} \approx \frac{5.1475}{2.0794}\approx 2.4754$

Rond to one decimal place.

2.5

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Example Question #41 : Understanding Logarithms

Solve for in the equation:

-log(x)=4.2

Possible Answers:

$x=e^{-4.2}$

$x=10^{-4.2}$

$x=10^{-log(4.2)}$

$x=e^{4.2}$

Correct answer:

$x=10^{-4.2}$

Explanation:

This question tests your understanding of log functions.

log(x)=y can be converted to the form $x=10^{y}$ .

In this problem, make sure to divide both sides by in order to put it in the above form, where log(x)=-4.2 . Remember $log(x)=log_{10}(x)$ .

Therefore,

$log_{10}(x)=-4.2 \rightarrow x=10^{-4.2}$

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Example Question #41 : Logarithms

Evaluate .

$5 ^{x} = 12$

Possible Answers:

$x= \frac{ \log 5}{12}$

$x= \log 7$

$x= \frac{ \log 12 }{5}$

$x=\frac{\log 12}{\log 5}$

$x=\frac{ \log 5}{\log 12}$

Correct answer:

$x=\frac{\log 12}{\log 5}$

Explanation:

Take the common logarithm of both sides, and take advantage of the property of the logarithm of a power:

$5 ^{x} = 12$

$\log 5 ^{x} = \log 12$

$x \log 5 = \log 12$

$\frac{x \log 5 }{\log 5 }=\frac{ \log 12}{\log 5}$

$x=\frac{ \log 12}{\log 5}$

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Example Question #41 : Logarithms

Which equation is equivalent to:

$\log _{4}\left( \frac{1}{2}\right)=x$

Possible Answers:

$(4)^{x}=\left(\frac{1}{2}\right)^{x}$

$4^{\frac{1}{2}}=x$

$\frac{1^{4}}{2}=x^{4}$

$4^{x}=\frac{1}{2}$

$\left(\frac{1}{2}\right)^{4}=x$

Correct answer:

$4^{x}=\frac{1}{2}$

Explanation:

$log{_{a}}^{b}=x$ , b>0

$\Rightarrow a^{x}=b$

So, ${{log_{4}}^{\frac{1}{2}}}=x\Rightarrow 4^{x}=\frac{1}{2}$

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Example Question #3 : Exponential And Logarithmic Functions

$Log_{b}x=y$

What is the inverse of the log function?

Possible Answers:

$y^{b}=x$

$x^{b}=y$

$b=x\cdot y$

$b^{x}=y$

$b^{y}=x$

Correct answer:

$b^{y}=x$

Explanation:

This is a general formula that you should memorize. The inverse of $Log_{b}x=y$ is $b^{y}=x$ . You can use this formula to change an equation from a log function to an exponential function.

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Example Question #42 : Understanding Logarithms

Solve for :

7^b=54

Round to the nearest hundredth.

Possible Answers:

3.21

2.05

Cannot be computed

1.45

0.49

Correct answer:

2.05

Explanation:

To solve this, you need to set up a logarithm. Our exponent is . The logarithm's base is . The value is the operand of the logarithm. Therefore, we can write an equation:

$\log_7{54} = b$

Now, you cannot do this on your calculator. Therefore, using the rule for converting logarithms, you need to change:

$\log_7{54}$

to...

$\frac{\log{54}}{\log{7}}$

You can put this into your calculator and get:

2.04993228926876 , or rounded, 2.05

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Algebra II : Logarithms

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #11 : Log Base 10

Example Question #11 : Log Base 10

Example Question #21 : Log Base 10

Example Question #22 : Log Base 10

Example Question #41 : Understanding Logarithms

Example Question #41 : Understanding Logarithms

Example Question #41 : Logarithms

Example Question #41 : Logarithms

Example Question #3 : Exponential And Logarithmic Functions

Example Question #42 : Understanding Logarithms

All Algebra II Resources

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