log equations?

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December 12th 2009, 07:44 PM
Barthayn

log equations?

I got this as a question:

Simplify. State any restrictions on the variables.

$2 log w + (3/2) log w + (1/2) log w^2$

For this question I did this:

$2 log w + (3/2) log w + (1/2) log w^2$
$2 log w + (3/2) log w + (1/2) x 2 log w$
$2 log w + (3/2) log w + log w$
$2 log w + (3/2) log w$
$(9/2) log w$

The $2 log w + log w$ is still 2 log because it is multiplying by one correct and $(1/2) log w^2$ is 1 log because you are multiplying $(1/2)$ by two to get one correct?
December 12th 2009, 08:01 PM
mathaddict

Quote:

Originally Posted by Barthayn

I got this as a question:

Simplify. State any restrictions on the variables.

$2 log w + (3/2) log w + (1/2) log w^2$

For this question I did this:

$2 log w + (3/2) log w + (1/2) log w^2$
$2 log w + (3/2) log w + (1/2) x 2 log w$
$2 log w + (3/2) log w + log w$
$2 log w + (3/2) log w$
$(9/2) log w$

The $2 log w + log w$ is still 2 log because it is multiplying by one correct and $(1/2) log w^2$ is 1 log because you are multiplying $(1/2)$ by two to get one correct?

HI

$\frac{9}{2}\log w$ is the correct simplication , you cant go beyond that .

For the log to be defined , $w\in Z^+$ or in other words w>0 , try log negative sth , ur calculator will return with math error .
December 12th 2009, 08:04 PM
Barthayn

I know it is. However, I do not understand why it is like that. Could you explain my steps, or explain how you would do it?
December 12th 2009, 08:06 PM
mathaddict

Quote:

Originally Posted by Barthayn

I know it is. However, I do not understand why it is like that. Could you explain my steps, or explain how you would do it?

you mean the steps to get (9/2)log w , or why w must be >0 ? which one ?
December 12th 2009, 08:11 PM
Barthayn

Quote:

Originally Posted by mathaddict

you mean the steps to get (9/2)log w , or why w must be >0 ? which one ?

I wish to understand how to do the steps to get the final answer. I understand that a negative log will give you an error. I am confused with the 1/2 log w^2
December 12th 2009, 08:17 PM
mathaddict

Quote:

Originally Posted by Barthayn

The $2 log w + log w$ is still 2 log because it is multiplying by one correct and $(1/2) log w^2$ is 1 log (correct)because you are multiplying $(1/2)$ by two to get one correct?

$2\log w+\log w=3\log w$

this part ?
December 12th 2009, 08:20 PM
Barthayn

ahh. I see my mistake. I guess I looked at my steps incorrectly. I see I did everything correct. Thank you for your help.