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Algebra II : Multiplying and Dividing Logarithms

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Example Questions

Example Question #1 : Multiplying And Dividing Logarithms

Rewrite the following logarithmic expression in expanded form (i.e. as a sum and/or difference):

$ln \left(\frac{x^{3}y^{-2}}{2z^{4}}\right)$

Possible Answers:

3ln(x) -2ln(y) -8ln(z)

3ln(x) -2ln(y) + ln(2) + 4ln(z)

3ln(x) -2ln(y) - ln(2) - 4 ln(z)

3ln(x) + 2ln(y) - ln(2) - 4ln(z)

3ln(x) - 2ln(y) -ln(2) + 4ln(z)

Correct answer:

3ln(x) -2ln(y) - ln(2) - 4 ln(z)

Explanation:

By logarithmic properties:

$ln(x^{4}) = 4ln (x)$ ;

$ln(y^{-2})= -2ln(y)$

$\frac{1}{2z^{4}} = -ln(2) - 4ln(z)$

Combining these three terms gives the correct answer:

3ln(x) -2ln(y) - ln(2) - 4 ln(z)

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Example Question #2 : Multiplying And Dividing Logarithms

Which of the following is equivalent to

$\log(4000)$ ?

Possible Answers:

$10 + \log4$

2000

$3\log4$

$3 + \log4$

Correct answer:

$3 + \log4$

Explanation:

Recall that log implies base if not indicated.Then, we break up $4000 = 4\cdot 1000$ . Thus, we have $\log(4 \cdot 1000)$ .

Our log rules indicate that

$\log(ab) = \log(a) + \log(b)$ .

So we are really interested in,

$\log(4) + \log(1000)$ .

Since we are interested in log base , we can solve $\log(1000)$ without a calculator.

We know that 1000=10^3 , and thus by the definition of log we have that $\log(1000) = 3$ .

Therefore, we have $\log(4) + 3$ .

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Example Question #3 : Multiplying And Dividing Logarithms

Find the value of the Logarithmic Expression.

$\log_{3} \frac{1}{81}$

Possible Answers:

Correct answer:

Explanation:

Use the change of base formula to solve this equation.

$Log_{_{3}} \frac{1}{81}$

$y = Log_{3} \frac{1}{81}$

$3^{y} = \frac{1}{81}$

$3^{^{y}} = 3^{-4}$

y = -4

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Example Question #4 : Multiplying And Dividing Logarithms

What is another way of expressing the following?

$\small 3\log(2)$

Possible Answers:

$\small \small \log (\frac{2}{3})$

$\small \log (6)$

$\small \log (9)$

$\small \log (8)$

$\small \log (5)$

Correct answer:

$\small \log (8)$

Explanation:

Use the rule

$\small a\times \log (b)= \log (b^{a})$

$\small \log (2^{3})=\log (8)$

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Example Question #5 : Multiplying And Dividing Logarithms

Expand this logarithm: $\log_{10}8x^4y$

Possible Answers:

$\log_{10}8y+4\log_{10}x$

$\log_{10}y+\log_{10}x^4 +8$

$\log_{10}8+\log_{10}x^4+\log_{10}y$

$\log_{10}(8+y+x^4)$

$\log_{10}8+4\log_{10}x+\log_{10}y$

Correct answer:

$\log_{10}8+4\log_{10}x+\log_{10}y$

Explanation:

In order to solve this problem you must understand the product property of logarithms $\log_{10}uv=\log_{10}u+\log_{10}v$ and the power property of logarithms $\log_{10}u^n=n\log_{10}u$ . Note that these apply to logs of all bases not just base 10.

$\log_{10}8x^4y$

log of multiple terms is the log of each individual one:

$\log_{10}8+\log_{10}x^4+\log_{10}y$

now use the power property to move the exponent over:

$\log_{10}8+4\log_{10}x+\log_{10}y$

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Example Question #6 : Multiplying And Dividing Logarithms

Which of the following is equivalent to $log(\frac{a^2b^3}{x^{-3}})$ ?

Possible Answers:

2log(a)+log(b-x)^3

log(2a)+log(3b)+log(3x)

2log(a)+3log(b-x)

2log(a)+3log(b)-3log(x)

2log(a)+3log(b)+3log(x)

Correct answer:

2log(a)+3log(b)+3log(x)

Explanation:

We can rewrite the terms of the inner quantity. Change the negative exponent into a fraction.

$x^{-3} = \frac{1}{x^3}$

$\frac{a^2b^3}{x^{-3}} = a^2b^3 \div \frac{1}{x^3}= a^2b^3 \times \frac{x^3}{1} =a^2b^3x^3$

This means that:

$log(\frac{a^2b^3}{x^{-3}}) = log(a^2b^3x^3)$

Split up these logarithms by addition.

log(a^2b^3x^3) = log(a^2)+log(b^3)+log(x^3)

According to the log rules, the powers can be transferred in front of the logs as coefficients.

The answer is: 2log(a)+3log(b)+3log(x)

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Example Question #7 : Multiplying And Dividing Logarithms

Many textbooks use the following convention for logarithms:

$log(x) = log_{10}(x)$

ln(x) = log_e(x)

lg(x)=log_2(x)

Solve:

$lg(\frac{4096}{16384})$

Possible Answers:

Correct answer:

Explanation:

Remembering the rules for logarithms, we know that $log_a(\frac{b}{c}) = log_ab - log_ac$ .

This tells us that $lg(\frac{4096}{16384})=lg(4096)-lg(16384)$ .

This becomes 12 - 14 , which is .

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Algebra II : Multiplying and Dividing Logarithms

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #1 : Multiplying And Dividing Logarithms

Example Question #2 : Multiplying And Dividing Logarithms

Example Question #3 : Multiplying And Dividing Logarithms

Example Question #4 : Multiplying And Dividing Logarithms

Example Question #5 : Multiplying And Dividing Logarithms

Example Question #6 : Multiplying And Dividing Logarithms

Example Question #7 : Multiplying And Dividing Logarithms

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