# Math Help - Condensing Logs :(

1. ## Condensing Logs :(

So i have been working on this problem for about 45 minutes now and cant really seem to be going anywhere. I ended up getting different answers each time and have no idea what im doing wrong. Can you guys help please?

Condense: 4(logx + 3log(x+2) + logy) - 2(log15 + log2 + logy) - 1/2log4 + 2/3log27-(4log 5 + 4log x + 2log y)-11log(x+2)

2. Originally Posted by layneek
So i have been working on this problem for about 45 minutes now and cant really seem to be going anywhere. I ended up getting different answers each time and have no idea what im doing wrong. Can you guys help please?

Condense: 4(logx + 3log(x+2) + logy) - 2(log15 + log2 + logy) - 1/2log4 + 2/3log27-(4log 5 + 4log x + 2log y)-11log(x+2)
You have two rules to follow:
$log(a) + log(b) = log(ab)$

and
$a \cdot log(b) = log(b^a)$

$4(log(x) + 3log(x+2) + log(y)) = 4(log[x(x + 2)]y) = log \{ ( [ x(x + 2) ] y)^4 \} = log(x^4(x + 2)^4 y^4)$

Now do this with your other terms.

-Dan

3. Hello, layneek!

$\text{Condense:}$

$4\bigg[\log x + 3\log(x+2) + \log y\bigg] - 2\bigg[\log15 + \log2 + \log y\bigg]$

. . . $- \frac{1}{2}\log4 + \frac{2}{3}\log27-\bigg[4\log 5 + 4\log x + 2\log y\bigg]-11\log(x+2)$

Clear parentheses:

$4\log x + 12\log(x+2) + 4\log y - 2\log15 - 2\log 2 - 2 \log y$
. . . . $-\frac{1}{2}\log4 + \frac{2}{3}\log27 - 4\log5 - 4\log x - 2\log y - 11\log(x+2)$

. . $=\;\log(x+2) + \frac{2}{3}\log27 -2\log15 - 2\log2 -\frac{1}{2}\log4 - 4\log5$

. . $=\;\log(x+2) + \log\left(27^{\frac{2}{3}}\right) - \log\left(15^2\right) - \log(2^2) - \log\left(4^{\frac{1}{2}}\right) - \log(5^4)$

. . $=\;\bigg[\log(x+2) + \log 9\bigg] - \bigg[\log225 + \log 4 + \log 2 + \log625\bigg]$

. . $=\;\log9(x+2) - \log(225\cdot4\cdot2\cdot625)$

. . $=\;\log9(x+2) - \log(1,\!125,\!000)$

. . $=\;\log\dfrac{9(x+2)}{1,125,000}$