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已知log2x+logx8=4,求x
人气:366 ℃ 时间:2020-02-02 10:14:11
解答
换底:
logx^8=(log2^8)/log2^x=3/log2^x
原方程变为:
log2^x+3/log2^x=4
(log2^x)^2-4log2^x+3=0
(log2^x-1)(log2^x-3)=0
log2^x=1或log2^x=3
故x=1或x=8
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