# Can someone tell what I'm doing wrong with this logerithm?

• November 27th 2007, 12:50 PM
Darkhrse99
Can someone tell what I'm doing wrong with this logerithm?
The problem is
Log6^18- Log 6^3.

When you have subtraction in a log the properties say divide. So I would assume I would divide 18/3 which is 6. But the answer to the problem is 1.

They both have a base of 6 so I know i don't have to change that. On my help cd that came with the text book says divide M/N which is what I have done.

Thanks

Jason
• November 27th 2007, 01:23 PM
Darkhrse99
Is this the reason? When you divide 18/3 =6 and the base is also 6 means if the base is 6 and the power is 6 then it= 1?
• November 27th 2007, 03:40 PM
Plato
This is not an answer. It is a complaint.
Why do you not learn to use notation?
I do not know what you are asking.
• November 27th 2007, 03:43 PM
shilz222
$\log_{6}18 - \log_{6}3 = \log_{6}6 = 1$.
• November 27th 2007, 04:01 PM
Plato
Quote:

Originally Posted by shilz222
$\log_{6}18 - \log_{6}3 = \log_{6}6 = 1$.

Why do you think that the question meant that?
• November 27th 2007, 04:32 PM
topsquark
Quote:

Originally Posted by Darkhrse99
Log6^18- Log 6^3.

I'm with Plato. I would have thought that this problem was $log_{10}(6^{18}) - log_{10}(6^3)$
which is decidedly not equal to 1.

@ Darkhrse99:
To type out $log_b(a)$ type in:
[ math ] <-- Without the spaces
log_{b}(a)
[ /math ] <-- Again, without the spaces.

I realize that learning LaTeX should not really be necessary to ask questions here, but I can't think of a good way to use the ASCII codes to do logarithms.

-Dan
• November 27th 2007, 06:06 PM
Darkhrse99
Quote:

Originally Posted by shilz222
$\log_{6}18 - \log_{6}3 = \log_{6}6 = 1$.

This is the correct question. I couldn't gt the little 6 in the equation right so [tex] formula didn't look right. Sorry I couldn't type it more clearly. On the other hand, I did get it right. :D