# Condensing Fractional Logarithms Does the following condense to the following: log_2 z+(log_2 x)/2+(log_2 y)/2=log_2 (zsqrt(x) sqrt(y)) or to log_2(z sqrt(xy)) ?

Condensing Fractional Logarithms
Does the following condense to the following:
${\mathrm{log}}_{2}z+\left({\mathrm{log}}_{2}x\right)/2+\left({\mathrm{log}}_{2}y\right)/2={\mathrm{log}}_{2}\left(z\sqrt{x}\sqrt{y}\right)$ or to ${\mathrm{log}}_{2}\left(z\sqrt{xy}\right)$?
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optativaspv
${\mathrm{log}}_{2}z+\frac{{\mathrm{log}}_{2}x}{2}+\frac{{\mathrm{log}}_{2}y}{2}={\mathrm{log}}_{2}z+{\mathrm{log}}_{2}{x}^{\frac{1}{2}}+{\mathrm{log}}_{2}{y}^{\frac{1}{2}}={\mathrm{log}}_{2}\left(z\cdot \sqrt{x}\cdot \sqrt{y}\right)={\mathrm{log}}_{2}\left(z\cdot \sqrt{x\cdot y}\right)$
We used the identities:
$\mathrm{log}a+\mathrm{log}b=\mathrm{log}\left(a\cdot b\right)$
$x\cdot \mathrm{log}a=\mathrm{log}{a}^{x}$
and
$\sqrt{a}\cdot \sqrt{b}=\sqrt{a\cdot b}$
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opositor5t
Both are the same, since $\sqrt{xy}=\sqrt{x}\cdot \sqrt{y}$