# combining logs

• January 26th 2010, 03:45 PM
ryan18
combining logs
Express the given quantity as a single logarithm. ln(a + b) + ln(a - b) - 9 ln(c)

Here is what I did:

ln a + ln b + ln a - ln b - 9 ln c
The ln b's cancel.

ln a + ln a - 9 ln c
Then combine to one logarithm.

ln(a^2 - c^9)

The outcome I got was incorrect. I was just wondering if you could point me in the right direction or point out my flaws. Thanks!
• January 26th 2010, 03:48 PM
skeeter
Quote:

Originally Posted by ryan18
Express the given quantity as a single logarithm. ln(a + b) + ln(a - b) - 9 ln(c)

Here is what I did:

ln a + ln b + ln a - ln b - 9 ln c
The ln b's cancel.

ln a + ln a - 9 ln c
Then combine to one logarithm.

ln(a^2 - c^9)

The outcome I got was incorrect. I was just wondering if you could point me in the right direction or point out my flaws. Thanks!

$\ln(a+b) \ne \ln{a} + \ln{b}$

$\ln(a \cdot b) = \ln{a} + \ln{b}$

I recommend that you review the properties of logarithms.
• January 26th 2010, 03:57 PM
ryan18
WOW. I dont know how I convinced myself of that!

Ok I got it now.

ln((a+b)(a-b)) - ln c^9
Factor out.

ln(a^2 - b^2) - ln c^9

ln((a^2 - b^2)/c^9)

:D sorry.
• January 26th 2010, 04:07 PM
pencil09
Quote:

Originally Posted by ryan18
Express the given quantity as a single logarithm. ln(a + b) + ln(a - b) - 9 ln(c)

Here is what I did:

ln a + ln b + ln a - ln b - 9 ln c
The ln b's cancel.

ln a + ln a - 9 ln c
Then combine to one logarithm.

ln(a^2 - c^9)

The outcome I got was incorrect. I was just wondering if you could point me in the right direction or point out my flaws. Thanks!

remember laws of logarithm...
ln (a b)= ln a+ ln b
ln $\frac{a}{b}$= ln a- ln b
so,
ln a + ln b + ln a - ln b - 9 ln c= ln $\frac{a b a}{b c^9}$