已知log2x+logx8=4,求x
问题描述：
已知log2x+logx8=4,求x
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