Question

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

Answer 1

分析 设log₈₁x=a，log₆₄y=b．可得$\left\{\begin{array}{l}{a+b=4}\\{\frac{1}{a}-\frac{1}{b}=1}\end{array}\right.$，化为a²-6a+4=0，利用根与系数的关系可得：x₁x₂=81⁶，同理可得y₁y₂=64²．代入即可得出．

解答 解：设log₈₁x=a，log₆₄y=b．
则$\left\{\begin{array}{l}{a+b=4}\\{\frac{1}{a}-\frac{1}{b}=1}\end{array}\right.$，化为a²-6a+4=0，
∴a₁+a₂=log₈₁（x₁x₂）=6，∴x₁x₂=81⁶，
同理可得y₁y₂=64²．
∴log₁₈（x₁x₂y₁y₂）=$lo{g}_{18}（8{1}^{6}×6{4}^{2}）$=12．
故答案为：12．

点评 本题考查了对数的运算法则、指数式与对数式的互化、换元法，考查了推理能力与计算能力，属于基础题．