高中数学题。求解。，解方程。
\u9ad8\u4e2d\u6570\u5b66\u8fd9\u4e2a\u65b9\u7a0b\u600e\u4e48\u89e3\uff1f\u89e3\uff1a
\u4ee4f(x)=lnx+x-1
\u5bf9\u6570\u6709\u610f\u4e49\uff0c\u771f\u6570>0\uff0cx>0\uff0c\u51fd\u6570\u7684\u5b9a\u4e49\u57df\u4e3a(0\uff0c+\u221e)
f'(x)=1/x +1
x>0\uff0c1/x>0\uff0c\u53c81>0\uff0c\u56e0\u6b641/x +1>0
f'(x)>0\uff0c\u51fd\u6570\u5728(0\uff0c+\u221e)\u4e0a\u5355\u8c03\u9012\u589e\uff0c\u81f3\u591a\u6709\u4e00\u4e2a\u96f6\u70b9\u3002
x=1\u65f6\uff0cf(x)=ln1 +1-1=0\uff0c\u51fd\u6570\u6709\u552f\u4e00\u96f6\u70b9x=1
\u65b9\u7a0blnx +x-1=0\u6709\u552f\u4e00\u5b9e\u6570\u89e3x=1
(1-x)*ln3 = ln2
那么：
1-x = ln2/ln3
所以：
x = 1 - ln2/ln3
解：原方程两边取对数得
1-n=log3 2(2是上标，和3的平方差别多写法）
则n=1-log3 2
请采纳
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