log equations?

• Dec 12th 2009, 07:44 PM
Barthayn
log equations?
I got this as a question:

Simplify. State any restrictions on the variables.

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

For this question I did this:

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

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

Originally Posted by Barthayn
I got this as a question:

Simplify. State any restrictions on the variables.

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

For this question I did this:

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

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

HI

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

For the log to be defined , $\displaystyle w\in Z^+$ or in other words w>0 , try log negative sth , ur calculator will return with math error .
• Dec 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?
• Dec 12th 2009, 08:06 PM
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 ?
• Dec 12th 2009, 08:11 PM
Barthayn
Quote:

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
• Dec 12th 2009, 08:17 PM
The $\displaystyle 2 log w + log w$ is still 2 log because it is multiplying by one correct and $\displaystyle (1/2) log w^2$ is 1 log (correct)because you are multiplying $\displaystyle (1/2)$ by two to get one correct?
$\displaystyle 2\log w+\log w=3\log w$