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2^(2+log2 3)+3^(2%log3 9)

2log₃ 2-log₃(32/9)+log₃8-3^(2+log₃5) =log₃4-log₃(32/9)+log₃8-3^2*3^log₃5 =log₃(4÷32/9×8) -9*5 =log₃(4×9/32×8)-45 =log₃9-45 =2-45 =-43 最后的式子指数应该加括号,...

(log3 2+log9 2)*(log4 3+log8 3) =(lg2/lg3+lg2/lg9)*(lg3/lg4+lg3/lg8) =(lg2/lg3+lg2/2lg3)*(lg3/2lg2+lg3/3lg2 =3/2*lg2/lg3*5/6*lg3/lg2 =3/2*5/6=5/4

(log3∧2+log9∧2)×(log4∧3+log8∧3)=(log3∧2+log(3^2)∧2)×(log(2^2)∧3+log(2^3)∧3)=(log3∧2+1/2log3∧2)×(1/2log2∧3+1/3log2∧3)=3/2log3^2*5/6log2^3=5/4

我会,就是把4变为二的平方,所以就是(6log2+3log8) (2log3+4log3)再分别相乘,得出结果!

(log3 2+log9 2)*(log4 3+log8 3) =(lg2/lg3+lg2/lg9)*(lg3/lg4+lg3/lg8) =(lg2/lg3+lg2/2lg3)*(lg3/2lg2+lg3/3lg2 =3/2*lg2/lg3*5/6*lg3/lg2 =3/2*5/6=5/4

解答: 定义域是x∈[1,9]且x²∈[1,9] ∴ x∈【1,3】 ∵ y=[2+log3(x)]²+2+log3(x²) =[log3(x)]²+4log3(x)+4+2+2log3(x) =[log3(x)+3]²-3 ∵ x∈[1,3] ∴ t=log3(x)∈[0,1] ∴ y=(t+3)²-3 ∴ t=0时,即x=1时,y有最小值6 t=1...

5^log5(2) - 4^log2(3) + 3^log27(8) =2 - 2^2log2(3) + 27^[(1/3)log27(8)] =2 - 2^log2(9) + 27^[log27(2)] =2 - 9 + 2 = -5

2log3(2)-log3(32/9)+log3(8)-5^log5(3) =log3(4)-log3(32/9)+log3(8)-3 =log3(4*8)-log3(32/9)-3 =log3(32)-log3(32/9)-3 =log3(32/(32/9))-3 =log3(9)-3 =2-3 =-1

log2(9)=log2(3²)=2log2(3) ∴log2(9)×log3(2)=2log2(3)×log3(2) =2×1 =2 望采纳

解析 可以这么来做 log3 9^(-1) =-log3 9 =-2 希望对你有帮助 学习进步O(∩_∩)O谢谢

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