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Does the following condense to the following:
$\log_2z+(\log_2x)/2+(\log_2y)/2 = \log_2(z\sqrt{x}\sqrt{y})$ or to $\log_2(z\sqrt{xy})$ ?

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  • $\begingroup$ The second one is simpler, but they are the same technically. $\endgroup$ Aug 26, 2014 at 23:40
  • $\begingroup$ So the second one is technically the most condensed? $\endgroup$
    – user169562
    Aug 26, 2014 at 23:40
  • $\begingroup$ @Arithmetician Define what you mean by "condensed" - if this is a homework problem and you're just trying to write a "simple answer," decide what you want to define as condensed. $\endgroup$ Aug 26, 2014 at 23:42
  • $\begingroup$ Yes, the second one is considered simpler. $\endgroup$ Aug 26, 2014 at 23:43

2 Answers 2

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Both of your answers are correct!

$$\log_2 z+ \frac{\log_2 x}{2}+\frac{\log_2 y}{2}=\log_2 z+\log_2 x^{\frac{1}{2}}+\log_2 y ^{\frac{1}{2}}=\log_2 (z \cdot \sqrt{x} \cdot \sqrt{y})=\log_2 (z \cdot \sqrt{x \cdot y})$$

We used the identities:

$$\log a+ \log b=\log (a \cdot b) $$ $$ x \cdot \log a=\log a^x$$

and

$$\sqrt{a} \cdot \sqrt{b}=\sqrt{a \cdot b}$$

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Both are the same, since $ \sqrt {xy} = \sqrt {x} \cdot \sqrt y $.

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