(2/3)^x≤1/20 (已知:lg2=0.3010,lg3=0.4771)
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(2/3)^x≤1/20(已知:lg2=0.3010,lg3=0.4771)(2/3)^x≤1/20(已知:lg2=0.3010,lg3=0.4771)(2/3)^x≤1/20(已知:lg2=0.30
(2/3)^x≤1/20 (已知:lg2=0.3010,lg3=0.4771)
(2/3)^x≤1/20 (已知:lg2=0.3010,lg3=0.4771)
(2/3)^x≤1/20 (已知:lg2=0.3010,lg3=0.4771)
两边同时取lg,因为lg是增函数,所以不等式方向不变,也就是lg(2/3)^x≤lg(1/20)
x lg(2/3)≤lg(1/20)
x(lg2-lg3)≤lg1-lg20
其中lg1=0,lg20=lg2+lg10=1.3010
所以x≥1.3010/(0.4771-0.3010)
x≥7.3878
不等式两边取lg,则lg(2/3)^x<1/20得出x*lg(2/3)<-(lg2*10)推出x<(-lg2-lg10)/(lg2-lg3),最后x<=13010/1761
(2/3)^x≤1/20
x*lg(2/3)≤lg(1/20)
x*(lg2-lg3)≤-(lg2+1)
x*(0.3010-0.4771)≤-(0.3010+1)
-0.1761*x ≤-1.3010
x ≥7.3878
(2/3)^x≤1/20 所以x≤log(2/3)(1/20) log(2/3)(1/20)=lg(1/20)/lg(2/3)=
(lg1-lg20)/(lg2-lg3)=(lg1-lg2-lg10)/(lg2-lg3) lg2=0.3010,lg3=0.4771 lg1=0 lg10=1
所以原试=-1.3010/-0.7781=1.6720