Question

14．當x為何值時，$\sqrt{lg\sqrt{lg\sqrt{lg\sqrt{lg\sqrt{lg\sqrt{lgx}}}}}}$才有意義．

Answer 1

分析 根據(jù)偶次被開方數(shù)大于等于零、真數(shù)大于零列出不等式組，根據(jù)對數(shù)的運算求解．

解答 解：要使函數(shù)有意義，則$\left\{\begin{array}{l}{x＞0}\\{lg\sqrt{lg\sqrt{lg\sqrt{lg\sqrt{lg\sqrt{lgx}}}}}}\end{array}\right.≥0$，
所以$\left\{\begin{array}{l}{x＞0}\\{\sqrt{lg\sqrt{lg\sqrt{lg\sqrt{lg\sqrt{lgx}}}}}}\end{array}\right.≥1$，則$\left\{\begin{array}{l}{x＞0}\\{lg\sqrt{lg\sqrt{lg\sqrt{lg\sqrt{lgx}}}}}\end{array}\right.≥1$，即$\left\{\begin{array}{l}{x＞0}\\{\sqrt{lg\sqrt{lg\sqrt{lg\sqrt{lgx}}}}}\end{array}\right.≥10$，
依次得：$\left\{\begin{array}{l}{x＞0}\\{lg\sqrt{lg\sqrt{lg\sqrt{lgx}}}}\end{array}\right.≥100$，…，
解得$x≥1{0}^{1{{0}^{2}}^{1{{0}^{2}}^{1{{0}^{2}}^{1{0}^{2}}}}}$，
所以函數(shù)的定義域是[$1{0}^{1{{0}^{2}}^{1{{0}^{2}}^{1{{0}^{2}}^{1{0}^{2}}}}}$，+∞）．

點評 本題考查對數(shù)函數(shù)的定義域，以及對數(shù)的運算，屬于中檔題．