Answer
$\frac{1}{2} + \frac{1}{2}\ln{x}$
Work Step by Step
Note that $\sqrt{ex} = (ex)^{\frac{1}{2}}$ therefore the expression above is equivalent to:
$=\ln{(ex)^{\frac{1}{2}}}$
RECALL:
(1) $\ln{(b^c)}=c \cdot \ln{b}$
(2) $\ln{(xy)} = \ln{x} + \ln{y}$
(3) $\ln{(\frac{x}{y})}=\ln{x} - \ln{y}$
Use rule (1) above to obtain:
$=\frac{1}{2}\ln{(ex)}$
Use rule (2) above to obtain:
$=\frac{1}{2}(\ln{e} + \ln{x})
\\=\frac{1}{2}\ln{e} + \frac{1}{2}\ln{x}$
Use the rule $\ln{e} = 1$ to obtain:
$=\frac{1}{2} \cdot 1 + \frac{1}{2}\ln{x}
\\=\frac{1}{2} + \frac{1}{2}\ln{x}$
