Multiplying and Dividing Logarithms - Algebra 2

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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}$}\right)$

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By logarithmic properties:

;

Combining these three terms gives the correct answer:

Many textbooks use the following convention for logarithms:

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

Solve:

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

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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 .

Find the value of the Logarithmic Expression.

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Use the change of base formula to solve this equation.

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

Which of the following is equivalent to

$\log(4000)$ ?

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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 $\log(4) + 3$ .

What is another way of expressing the following?

$\small 3\log(2)$

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Use the rule

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

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

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

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In order to solve this problem you must understand the product property of logarithms 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:

now use the power property to move the exponent over:

Which of the following is equivalent to ?

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

This means that:

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:

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}$}\right)$

Tap to see back →

By logarithmic properties:

;

Combining these three terms gives the correct answer:

Many textbooks use the following convention for logarithms:

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

Solve:

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

Tap to see back →

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 .

Find the value of the Logarithmic Expression.

Tap to see back →

Use the change of base formula to solve this equation.

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

Which of the following is equivalent to

$\log(4000)$ ?

Tap to see back →

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 $\log(4) + 3$ .

What is another way of expressing the following?

$\small 3\log(2)$

Tap to see back →

Use the rule

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

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

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

Tap to see back →

In order to solve this problem you must understand the product property of logarithms 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:

now use the power property to move the exponent over:

Which of the following is equivalent to ?

Tap to see back →

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

This means that:

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:

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}$}\right)$

Tap to see back →

By logarithmic properties:

;

Combining these three terms gives the correct answer:

Many textbooks use the following convention for logarithms:

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

Solve:

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

Tap to see back →

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 .

Find the value of the Logarithmic Expression.

Tap to see back →

Use the change of base formula to solve this equation.

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

Which of the following is equivalent to

$\log(4000)$ ?

Tap to see back →

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 $\log(4) + 3$ .

What is another way of expressing the following?

$\small 3\log(2)$

Tap to see back →

Use the rule

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

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

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

Tap to see back →

In order to solve this problem you must understand the product property of logarithms 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:

now use the power property to move the exponent over: