@wjl371116 第一步的x=log‹7/25›(34/35),这个我能看懂,=(lg34-lg35)/(lg7-lg25)是为什么呀?是用什么法则啊?
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![@wjl371116 第一步的x=log‹7/25›(34/35),这个我能看懂,=(lg34-lg35)/(lg7-lg25)是为什么呀?是用什么法则啊?](/uploads/image/z/13633369-25-9.jpg?t=%40wjl371116+%E7%AC%AC%E4%B8%80%E6%AD%A5%E7%9A%84x%3Dlog%26%238249%3B7%2F25%26%238250%3B%2834%2F35%29%EF%BC%8C%E8%BF%99%E4%B8%AA%E6%88%91%E8%83%BD%E7%9C%8B%E6%87%82%EF%BC%8C%3D%28lg34-lg35%29%2F%28lg7-lg25%29%E6%98%AF%E4%B8%BA%E4%BB%80%E4%B9%88%E5%91%80%EF%BC%9F%E6%98%AF%E7%94%A8%E4%BB%80%E4%B9%88%E6%B3%95%E5%88%99%E5%95%8A%EF%BC%9F)
@wjl371116 第一步的x=log‹7/25›(34/35),这个我能看懂,=(lg34-lg35)/(lg7-lg25)是为什么呀?是用什么法则啊?
@wjl371116 第一步的x=log‹7/25›(34/35),这个我能看懂,=(lg34-lg35)/(lg7-lg25)是为什么呀?是用什么法则啊?
@wjl371116 第一步的x=log‹7/25›(34/35),这个我能看懂,=(lg34-lg35)/(lg7-lg25)是为什么呀?是用什么法则啊?
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.
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) / (...
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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) / (7^x) = > 35/34 = ((5^2) / (7))^x
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看不到题目啊?