# Thread: World of Logs Rules

1. ## World of Logs Rules

Write as the sum and /or difference of logs. Express powers as factors.

16. In (x+8)(x-9)^3/2 / ((x-4)^4 , x>9

A. 3/2In(x+8)+3/2In(x-9)-6In(x-4)
B. In(x+8)+In(x-9)+In3-12In(x-4)-In2
C.3/2In(x^2+17x-72)-6In(x-4)
D.3In(x+8)-2In(x-9)-6In(x-4)

2. Originally Posted by magentarita
Write as the sum and /or difference of logs. Express powers as factors.

16. In (x+8)(x-9)^3/2 / ((x-4)^4 , x>9

A. 3/2In(x+8)+3/2In(x-9)-6In(x-4)
B. In(x+8)+In(x-9)+In3-12In(x-4)-In2
C.3/2In(x^2+17x-72)-6In(x-4)
D.3In(x+8)-2In(x-9)-6In(x-4)
$\displaystyle \ln{|\frac{(x+8)(x+9)^{\frac{3}{2}}}{(x-4)^4}|}$

Is that what it is supposed to look like?

If so:

$\displaystyle \ln{|(x+8)(x+9)^{\frac{3}{2}}|}-\ln{|(x-4)^4|}$

$\displaystyle \ln{|(x+8)|}+\ln{|(x+9)^{\frac{3}{2}}|}-\ln{|(x-4)^4|}$

$\displaystyle \ln{|(x+8)|}+\frac{3}{2}\ln{|(x+9)|}-4\ln{|(x-4)|}$

Originally Posted by Mush
$\displaystyle \ln{|\frac{(x+8)(x+9)^{\frac{3}{2}}}{(x-4)^4}|}$

Is that what it is supposed to look like?

If so:

$\displaystyle \ln{|(x+8)(x+9)^{\frac{3}{2}}|}-\ln{|(x-4)^4|}$

$\displaystyle \ln{|(x+8)|}+\ln{|(x+9)^{\frac{3}{2}}|}-\ln{|(x-4)^4|}$

$\displaystyle \ln{|(x+8)|}+\frac{3}{2}\ln{|(x+9)|}-4\ln{|(x-4)|}$