# Need logarithm work Checked

• Apr 22nd 2013, 11:19 AM
petenice
Need logarithm work Checked
Hi. I just wanted to see if someone could check my work over. The only one im not so sure of is number 4 ( or bottom right) Thank you.
• Apr 22nd 2013, 11:37 AM
dokrbb
Re: Need logarithm work Checked
Quote:

Originally Posted by petenice
Hi. I just wanted to see if someone could check my work over. The only one im not so sure of is number 4 ( or bottom right) Thank you.

concerning the last one - (a+b)^2 expands as a^2 + 2ab + b^2, but I'm not sure you have to go farther than the third step, though,

dokrbb
• Apr 22nd 2013, 11:47 AM
ebaines
Re: Need logarithm work Checked
You were fine up to

$\ln \frac {a-b}{(a+b)^2}$

But then with the next step you have $(a+b)^2 = a^2+ b^2$, which is incorrect. If you multiply it out you get:

$(a+b)^2 = (a+b)(a+b) = a^2 + 2ab + b^2$.

In addition the step after that you go from $\frac {a+b}{a^2 + b^2}$ to $\frac {1 + b}{a + b^2}$. This too is incorrect - if you were trying to divide both numerator and denominator by 'a' you would get $\frac {1 + b/a}{a+ b^2/a}$.

Also, although you didn't ask - you have an error in the first one: check the sign of the $\log_a z^2$ term in the second line.
• Apr 22nd 2013, 12:48 PM
petenice
Re: Need logarithm work Checked
Ebaines - I actually meant for all of them to be checked so thank you for doing so. I think i have fixed the problem with number 4. Can you check again? I did it two different ways. I think they are both correct.

Thank you.
• Apr 22nd 2013, 01:19 PM
ebaines
Re: Need logarithm work Checked
That's better - good job!