Answer
$1 + \frac{1}{2}\log{x}$
Work Step by Step
Simplify the radical to obtain:
$=\sqrt{10^2\cdot x} = 10\sqrt{x}$
Thus, the given expression is equivalent to:
$=\log{(10\sqrt{x})}$
RECALL:
(1) $\log{(b^c)}=c \cdot \log{b}$
(2) $\log_b{(xy)} = \log_b{x} + \log_b{y}$
(3) $\log_b{(\frac{x}{y})}=\log_b{x} - \log_b{y}$
Use rule (2) above to obtain:
$=\log{10}+\log{(\sqrt{x})}$
Note that $\sqrt{x} = x^{\frac{1}{2}}$ therefore the expression above is equivalent to:
$=\log{10} + \log{(x^{\frac{1}{2}})}$
Use rule (1) above to obtain:
$=\log{10} + \frac{1}{2}\log{x}$
Use the rule $\log{10} = 1$ to obtain:
$=1 + \frac{1}{2}\log{x}$
