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(log2 3+log4 27)(log3 4+log9 8)

解:【log2(3)+log4(9)+log8(27)+log16(81)+log32(243)】-5log2(3/2) =【log2(3)+log2^2(3^2)+log2^3(3^3)+log2^4(3^4)+log2^5(3^5)】-log2【(3/2)^5】 =【log2(3)+log2(3)+log2(3)+log2(3)+log2(3)】-log2【...

1、log2 3*log3 4*log4 5*log5 2 =lg2/lg3xlg3/lg4xlg4/lg5xlg5/lg2 (换底公式) =1 (约分) 2.(log4 3+log8 3)*(log3 2+log9 2) 中间应为乘号 =(lg3/lg4+lg3/lg8)*(lg2/lg3+lg2/lg9) =(lg3/2lg2+lg3/3lg2)*(lg2/lg3+lg2/2lg3) =[(1/2+1/3)(lg...

我来告诉你 我也是刚刚自己算出来的。 =Lg3/lg2 Lg8/lg3 +Lg9/lg4 Lg4/lg3 =Lg3/lg2 3Lg3/lg3 +2Lg3/lg4 Lg4/lg3 =3+2=5

(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

用换底公式 原式=(lg3/lg4+lg3/lg8)(lg2/lg3+lg2/lg9) =(lg3/2lg2+lg3/3lg2)(lg2/lg3+lg2/2lg3) =[(1/2+1/3)(lg3/lg2)][(1+1/2)(lg2/lg3)] =(1/2+1/3)(1+1/2) =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!

最终答案:71/12

(log4³+log8³)(log3²+log9²) =(lg3/lg4+lg3/lg8)×(lg2/lg3+lg2/lg9) =[lg3/(2lg2) + lg3/(3lg2)]×[lg2/lg3 + lg2/(2lg3)] =[(lg3/lg2)(1/2+1/3)]×[(lg2/lg3)(1+1/2)] =(1/2+1/3)(1+1/2) =5/6×3/2 =5/4 希望我的回答对你有帮...

因为这里书写不便,故将我的答案做成图像贴于下方,谨供楼主参考(若图像显示过小,点击图片可以放大)

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

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