Expressing a logarithmic equation

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Homework Statement



http://img142.imageshack.us/img142/8564/76203773.jpg

Homework Equations



I need to somehow express the above equation in terms of logax, logay,and logaz.

The Attempt at a Solution



I got the answer loga * [ |x|*sqrt (x)] + loga * [|y|] - loga * [z]

But I'd just like to get your guys' opinion, because I have the feeling I'm wrong, lol.
 
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How'd you get those answers? Did you use these laws?

log_a xy= log_a x + log_a y

log_ax^r=rlog_a x

log_a \frac{x}{y} = log_a x - log_a y
 
And your expression (NOT an equation) can be written as
\log_a (\frac{x^3 y^2}{z})^{1/2}
so you can immediately use the second law that rock.freak667 listed, and continue from there.
 
Or, not necessary but might help,
log_a\left(x^{3/2}yz^{-1/2}\right)
 
Would http://img23.imageshack.us/img23/7739/97409454.jpg be my final answer?
 
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No, for two reasons.
  1. It's wrong. The 1/2 factor should multiply all three terms.
  2. It can be simplified more in the x and y terms.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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