Question

14．已知方程組$\left\{\begin{array}{l}{lo{g}_{81}x+lo{g}_{64}y=4}\\{lo{g}_{x}81-lo{g}_{y}64=1}\end{array}\right.$的解為$\left\{\begin{array}{l}{x={x}_{1}}\\{y={y}_{1}}\end{array}\right.$和$\left\{\begin{array}{l}{x={x}_{2}}\\{y={y}_{2}}\end{array}\right.$，則log₁₈（x₁x₂y₁y₂）=12．

Answer 1

分析 設(shè)log₈₁x=a，log₆₄y=b．可得$\left\{\begin{array}{l}{a+b=4}\\{\frac{1}{a}-\frac{1}=1}\end{array}\right.$，化為a²-6a+4=0，利用根與系數(shù)的關(guān)系可得：x₁x₂=81⁶，同理可得y₁y₂=64²．代入即可得出．

解答 解：設(shè)log₈₁x=a，log₆₄y=b．
則$\left\{\begin{array}{l}{a+b=4}\\{\frac{1}{a}-\frac{1}=1}\end{array}\right.$，化為a²-6a+4=0，
∴a₁+a₂=log₈₁（x₁x₂）=6，∴x₁x₂=81⁶，
同理可得y₁y₂=64²．
∴l(xiāng)og₁₈（x₁x₂y₁y₂）=$lo{g}_{18}（8{1}^{6}×6{4}^{2}）$=12．
故答案為：12．

點(diǎn)評(píng) 本題考查了對(duì)數(shù)的運(yùn)算法則、指數(shù)式與對(duì)數(shù)式的互化、換元法，考查了推理能力與計(jì)算能力，屬于基礎(chǔ)題．