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.

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- Apr 22nd 2013, 10:19 AMpeteniceNeed 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, 10:37 AMdokrbbRe: Need logarithm work Checked
- Apr 22nd 2013, 10:47 AMebainesRe: Need logarithm work Checked
You were fine up to

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

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

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

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

Also, although you didn't ask - you have an error in the first one: check the sign of the $\displaystyle \log_a z^2$ term in the second line. - Apr 22nd 2013, 11:48 AMpeteniceRe: 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, 12:19 PMebainesRe: Need logarithm work Checked
That's better - good job!