# laws of logarithms

• Dec 17th 2009, 01:40 AM
extraordinarymachine
laws of logarithms
this question is trickier than i'm used too...a little help please?

$\displaystyle log_{7}(49)(7) + log_{7}\sqrt{7}$
• Dec 17th 2009, 01:47 AM
Stroodle
Hi there.

$\displaystyle log_7 49=2$

so $\displaystyle 7log_7 49=7\times2$

and $\displaystyle log_7\sqrt{7}=\frac{1}{2}$

So $\displaystyle 7log_7 49+log_7 \sqrt{7}=14\frac{1}{2}$
• Dec 17th 2009, 02:00 AM
extraordinarymachine
Quote:

Originally Posted by Stroodle
Hi there.

$\displaystyle log_7 49=2$

so http://www.mathhelpforum.com/math-he...b48564eb-1.gif

http://www.mathhelpforum.com/math-he...91bd8763-1.gif

i don't understand this part, why did the 7 move to the front of the log?
• Dec 17th 2009, 02:13 AM
Stroodle
Because it's in a separate bracket to the 49, it means that the logarithm is multiplied by 7. I just put it in front because that's how it is conventionally written (at least in my texts).
• Dec 17th 2009, 02:13 AM
Raoh
Quote:

Originally Posted by Stroodle
Hi there.

$\displaystyle log_7 49=2$

so $\displaystyle 7log_7 49=7\times2$

and $\displaystyle log_7\sqrt{7}=\frac{1}{2}$

So $\displaystyle 7log_7 49+log_7 \sqrt{7}=14+\frac{1}{2}$

you forgot a sign.
• Dec 17th 2009, 02:28 AM
Stroodle
Quote:

Originally Posted by Raoh
you forgot a sign.

I don't get it?

$\displaystyle 14+\frac{1}{2}=\frac{29}{2}=14\frac{1}{2}$