# Can somebody complete these task?

• Dec 8th 2010, 06:51 AM
Antalya
I need somebody to complete this little task, I wouldn't ask if there was a chance for me to understand this, but I don't until somebody complete this task.

a) Use logarithm rules to simplify the expression.

$lg a^{3}-lg(a^{2}b^{4})+2lg\frac{b^{2}}{a}$

b) Find the equations of the two asymptotes of the function g, where

$g(x)=\frac{2x+1}{x-1}$

a) $-lga(=lg\frac{1}{a})$
b) Vertical asymptote: x = 1
Horizontal asymptote: y = -2

I need full calculation in order.
• Dec 8th 2010, 07:09 AM
Sudharaka
Quote:

Originally Posted by Antalya
I need somebody to complete this little task, I wouldn't ask if there was a chance for me to understand this, but I don't until somebody complete this task.

a) Use logarithm rules to simplify the expression.

$lg a^{3}-lg(a^{2}b^{4})+2lg\frac{b^{2}}{a}$

b) Find the equations of the two asymptotes of the function g, where

$g(x)=\frac{2x+1}{x-1}$

a) $-lga(=lg\frac{1}{a})$
b) Vertical asymptote: x = 1
Horizontal asymptote: y = -2

I need full calculation in order.

Dear Antalya,

To solve the first one you have to know that,

$log(A\times B)=logA+logB$

$log\left(\frac{A}{B}\right)=logA-logB$

$logA^n=nlogA$

Hope you would be able to continue.
• Dec 8th 2010, 07:15 AM
Antalya
We have began on a chapter that i does not understand a piece of, and i have serious problems with this chapter. I don't understand this. Can anybody solve this task so i maybe can get an understanding on this. I have mentioned this before in my first tread.
• Dec 8th 2010, 07:24 AM
HallsofIvy
If you will not try you will never do anything.

Sudharaka gave you the information you needed- the "laws of logarithms".
lg(ab)= lg(a)+ lg(b)
lg(a/b)= lg(a)- lg(b)
$lg(a^b)= b lg(a)$

Now apply those to your problem, one piece at a time:
$lg(a^3)=$ what? Which of those laws applies?

$lg(a^2b^4)$ requires you to use two of those "laws"
$lg(a^2b^4)= lg( )+ lg( )$. What goes into the parentheses on the right? Then finish it.

$lg(b^2/a)= lg( )- lg( )$. Again, what goes in the parentheses?
• Dec 8th 2010, 07:37 AM
Antalya
I understand what you want to forward to, but I just want to know this on my own initiative because we have calculators and Geogebra to like this. Can not you just be kind enough to figure out this task, it will greatly simplify the process when I see how you've figured out the task because I can then use the same initial manner in the future. I don't understand short versions of task, if you give me a complete answer to this task then I don't need to bother you anymore with more of these types of task.
• Dec 8th 2010, 12:34 PM
mr fantastic
Quote:

Originally Posted by Antalya
I understand what you want to forward to, but I just want to know this on my own initiative because we have calculators and Geogebra to like this. Can not you just be kind enough to figure out this task, it will greatly simplify the process when I see how you've figured out the task because I can then use the same initial manner in the future. I don't understand short versions of task, if you give me a complete answer to this task then I don't need to bother you anymore with more of these types of task.

You have been asked - twice - to show some effort. Until you do, there is nothing more can be done to help you. MHF does not just give out solutions, especially when no effort has been shown.