Multiplying and Dividing Logarithms

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Algebra II › Multiplying and Dividing Logarithms

Questions 1 - 7

1

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

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

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:

2

Which of the following is equivalent to

$\log(4000)$ ?

$3 + \log4$

$3\log4$

$10 + \log4$

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

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

3

Find the value of the Logarithmic Expression.

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

Explanation

Use the change of base formula to solve this equation.

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

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

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

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

4

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

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.

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

The answer is:

5

What is another way of expressing the following?

$\small 3\log(2)$

$\small \log (8)$

$\small \log (6)$

$\small \log (5)$

$\small \log (9)$

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

Explanation

Use the rule

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

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

6

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

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

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

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

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

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

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:

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

7

Many textbooks use the following convention for logarithms:

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

Solve:

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

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 , which is .

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