the question is...
$\displaystyle \frac{1}{2}log(\frac{\sqrt{x}}{xy})$
my attempt...
$\displaystyle \frac{1}{2}log\sqrt{x}-\frac{1}{2}log(xy)$
$\displaystyle log{x}-\frac{1}{2}log(xy)$
the bit what i was unsire about was the sqrt
many thanks
the question is...
$\displaystyle \frac{1}{2}log(\frac{\sqrt{x}}{xy})$
my attempt...
$\displaystyle \frac{1}{2}log\sqrt{x}-\frac{1}{2}log(xy)$
$\displaystyle log{x}-\frac{1}{2}log(xy)$
the bit what i was unsire about was the sqrt
many thanks
Let's split this up and simplify the
$\displaystyle \frac{1}{2}log(x^\frac{1}{2})$ first.
Do you recall that
$\displaystyle Log(a^b)=b\times log(a)$?
How would you apply that rule here?
Now let's look at:
$\displaystyle -\frac{1}{2}log(xy)$
$\displaystyle = -\frac{1}{2} [log(xy)]$
Now, to simplify this, why not just apply the
rule?
$\displaystyle
=\frac{1}{2}\times\frac{1}{2}Log(x)-\frac{1}{2}[Log(x)+Log(y)]
$
$\displaystyle =\frac{1}{4}Log(x)-\frac{1}{2}log(x)-\frac{1}{2}Log(y)$
$\displaystyle =\frac{1}{2}[\frac{1}{2}Log(x)-Log(x)-Log(y)]$
$\displaystyle =\frac{1}{2}[-\frac{1}{2}Log(x)-Log(y)]$
I think that this is correct, I'll check in a second. So you just made a sign error somewhere.
Edit: I've checked, and it seems that I am correct. I must apologize for the late reply - I was caught on my PC when I should have been asleep, so I was forced to turn it off abrutly.