25-26高一上·湖南衡阳期末·第12题

$\boldsymbol{\text{e}^2}$或$\boldsymbol{0}$

【题目】方程$x^{\ln6} + x^{\ln8} = x^{\ln10}$的实数解为__.

【答案】$\boldsymbol{\text{e}^2}$或$\boldsymbol{0}$

【解析】$x=0$时方程$x^{\ln6} + x^{\ln8} = x^{\ln10}$显然成立，若$x \neq 0$，则有$x>0$.

设$t = \ln x$，则$x = \text{e}^t$，

可得$x^{\ln6} = (\text{e}^t)^{\ln6} = (\text{e}^{\ln6})^t = 6^t$，

同理可得$x^{\ln8} = 8^t$，$x^{\ln10} = 10^t$，

则原方程可化为$6^t + 8^t = 10^t$.

等式$6^t + 8^t = 10^t$两边同时除以$10^t$可得

$\left(\frac{3}{5}\right)^t + \left(\frac{4}{5}\right)^t = 1$,

即$\left(\frac{3}{5}\right)^t + \left(\frac{4}{5}\right)^t – 1 = 0$.

设$f(t) = \left(\frac{3}{5}\right)^t + \left(\frac{4}{5}\right)^t – 1$，

因为函数$y=\left(\frac{3}{5}\right)^t$、$y=\left(\frac{4}{5}\right)^t – 1$在$\mathbf{R}$上均为减函数，

故函数$f(t) = \left(\frac{3}{5}\right)^t + \left(\frac{4}{5}\right)^t – 1$在$\mathbf{R}$上为减函数.

又因为$f(2) = \left(\frac{3}{5}\right)^2 + \left(\frac{4}{5}\right)^2 – 1 = 0$,

所以方程$6^t + 8^t = 10^t$有且只有一个实根$t=2$.

由$\ln x = 2$，解得$x = \text{e}^2$，

故方程$x^{\ln6} + x^{\ln8} = x^{\ln10}$的实数解为$\text{e}^2$或$0$.

故答案为：$\boldsymbol{\text{e}^2}$或$\boldsymbol{0}$.