c2 help - logs

In a geometric sequence you go to the next term by multiplying by the common ratio, and go to the previous term by dividing by the common ratio. So, if the sequence has first term $a_1$, then $a_2 = a_1r$, and $a_3 = a_2r = a_1r^2$.

Here, you're given the values of $a_1, a_2, a_3$ in terms of $k$, and you're told to show that something is equal to zero. Now, since we know that $a_2 = a_1r$ and $a_3 = a_2r$, we get the result that $r = \dfrac{a_2}{a_1} = \dfrac{a_3}{a_2}$ (which is incidentally also why it's called the "common ratio", because $\dfrac{a_{n+1}}{a_n} = r$ for any n, but that's outside of the scope of this question).

Since $r - r = 0$, we can replace the first r by $\dfrac{a_3}{a_2}$ and the second r by $\dfrac{a_2}{a_1}$ and set them equal to zero. Writing these in terms of $k$ gives $\dfrac{2k-15}{k} - \dfrac{k}{k+4} = 0$. Multiplying through by $k(k+4)$ gives $(2k-15)(k+4) - k^2 = 0$... try and guess what you have to do now

You were the one willing to help me step by step, so thought you might be able to help me with 8b) as well on the above linked paper.
Thanks in advance

Hey!

Its quite a long solution, so I took a picture of my worked solution (which is correct.) You can find it at:

http://img707.imageshack.us/img707/4513/img0409c.jpg

I apologise about my handwriting :L I remember when I did this paper, I rushed it :L
If you cant understand it still, please do ask me