express in terms log(x) log(y)

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March 6th 2010, 12:21 PM
decoy808

express in terms log(x) log(y)

the question is...

$\frac{1}{2}log(\frac{\sqrt{x}}{xy})$

my attempt...

$\frac{1}{2}log\sqrt{x}-\frac{1}{2}log(xy)$

$log{x}-\frac{1}{2}log(xy)$

the bit what i was unsire about was the sqrt

many thanks
March 6th 2010, 01:04 PM
Quacky

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

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

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

So to start, I would say:

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

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

How would you continue to simplify?
March 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?
March 6th 2010, 04:30 PM
Quacky

http://www.mathhelpforum.com/math-he...88ac28a5-1.gif

Let's split this up and simplify the

$\frac{1}{2}log(x^\frac{1}{2})$ first.

Do you recall that

$Log(a^b)=b\times log(a)$ ?

How would you apply that rule here?

Now let's look at:

$-\frac{1}{2}log(xy)$

$= -\frac{1}{2} [log(xy)]$

Now, to simplify this, why not just apply the

http://www.mathhelpforum.com/math-he...da03a94e-1.gif rule?
March 6th 2010, 04:56 PM
decoy808

$\frac{1}{2}(-\frac{1}{2}log(x)+log(y))$ ?
March 6th 2010, 05:03 PM
Quacky

Edit, nevermind, will edit in a sec.
March 7th 2010, 02:23 AM
Quacky

http://www.mathhelpforum.com/math-he...88ac28a5-1.gif

$<br /> =\frac{1}{2}\times\frac{1}{2}Log(x)-\frac{1}{2}[Log(x)+Log(y)]<br />$

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

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

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

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)
March 7th 2010, 03:40 AM
decoy808

thank you for your help(Cool)