Answer
$19+20 \cdot \ln{x}$
Work Step by Step
RECALL:
(1) $\log_b{(\frac{m}{n})}=\log_b{m} - \log_b{n}$
(2) $\log_b{(mn)}=\log_b{m} + \log_b{n}$
(3) $\log_b{(b^x)}=x$
(4) $\log_b{(m^n)}=n \cdot \log_b{m}$
(5) $\sqrt{a} = a^{\frac{1}{2}}$
With $b=e$, use rule (2) above to obtain:
$=\ln{(e^{19})}+\ln{(x^{20})}$
Use rule (3) above with $b=e$ to obtain:
$=19+\ln{(x^{20})}$
Use rule (4) above to obtain:
$=19+20 \cdot \ln{x}$