1. If $x, y>0, \log _{x} y+\log , x=\frac{10}{3}, x y=$ 144, then the value of $x+y$ is ( ).
(A) $20 \sqrt{3}$
(B) $26 \sqrt{3}$
(C) $24 \sqrt{3}$
(D) $10 \sqrt{3}$
$-、 1 . B$.
Since $\log _{x} y=\frac{1}{\log _{y} x}$, and $\log _{x} y+\log x=\frac{10}{3}$, thus $\log _{x} y=3$ or $\frac{1}{3}$.
Assume $\log _{x} y=3$, then $y=x^{3}$.
Substitute into $x y=144$, solve to get $x=2 \sqrt{3}, y=24 \sqrt{3}$.
Therefore, $x+y=26 \sqrt{3}$.