this question is trickier than i'm used too...a little help please?

$\displaystyle log_{7}(49)(7) + log_{7}\sqrt{7}$

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- Dec 17th 2009, 01:40 AMextraordinarymachinelaws 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 AMStroodle
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 AMextraordinarymachine

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 AMStroodle
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 AMRaoh
- Dec 17th 2009, 02:28 AMStroodle