# express in terms log(x) log(y)

• Mar 6th 2010, 12:21 PM
decoy808
express in terms log(x) log(y)
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
• Mar 6th 2010, 01:04 PM
Quacky
http://www.mathhelpforum.com/math-he...b7510f6c-1.gif

Note that $\displaystyle \sqrt x = x^{\frac{1}{2}}$

Also note that $\displaystyle log(xy) = log(x)+log(y)$

So to start, I would say:

http://www.mathhelpforum.com/math-he...b7510f6c-1.gif

$\displaystyle =\frac{1}{2}log(x^\frac{1}{2})-\frac{1}{2}Log(xy)$

How would you continue to simplify?
• Mar 6th 2010, 03:40 PM
decoy808
im geting a bit confused with the simplifying.

do i remove the 1/2 by multipying by 2?

and im unsure how i would bring the x of the xy to the other side?
• Mar 6th 2010, 04:30 PM
Quacky
http://www.mathhelpforum.com/math-he...88ac28a5-1.gif

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

http://www.mathhelpforum.com/math-he...da03a94e-1.gif rule?
• Mar 6th 2010, 04:56 PM
decoy808
$\displaystyle \frac{1}{2}(-\frac{1}{2}log(x)+log(y))$ ?
• Mar 6th 2010, 05:03 PM
Quacky
Edit, nevermind, will edit in a sec.
• Mar 7th 2010, 02:23 AM
Quacky
http://www.mathhelpforum.com/math-he...88ac28a5-1.gif

$\displaystyle =\frac{1}{2}\times\frac{1}{2}Log(x)-\frac{1}{2}[Log(x)+Log(y)]$

$\displaystyle =\frac{1}{4}Log(x)-\frac{1}{2}log(x)-\frac{1}{2}Log(y)$

$\displaystyle =\frac{1}{2}[\frac{1}{2}Log(x)-Log(x)-Log(y)]$

$\displaystyle =\frac{1}{2}[-\frac{1}{2}Log(x)-Log(y)]$

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.(Giggle)
• Mar 7th 2010, 03:40 AM
decoy808