@wjl371116 第一步的x=log‹7/25›(34/35)，这个我能看懂，=(lg34-lg35)/(lg7-l

解方程
(1).5^(2x)-(7^x)-35[5^(2x)]+36(7^x)=0
-34[5^(2x)]+35(7^x)=0
-34(25^x)+35(7^x)=0
35(7^x)=34(25^x)
(7/5)x=34/35；故x=log₁.₄(34/35)=(lg34-lg35)/lg1.4=-0.086151348.
(2).e^(4t)-e^(2t)-6=0
设e^(2t)=u,则有u²-u-6=(u-3)(u+2)=0,故的u₁=3；u₂=-2(舍去)
由e^(2t)=3,得2t=lne,t=(1/2)ln3.

看不到题目啊？

ii) = (e^2t)^2 - e^2t - 6 = (e^2t - 3) (e^2t + 2) = 0;
=> e^2t - 3 = 0 => e^2t = 3 => 2t = ln3 = > t = ln3/2
i) -34(5^2x) + 35(7^x) = 0;
=> 35(7^x) = 34(5^2x) => 35/34 = (5^2x) / (...