这几道对数函数题求详细解答啊……9^1/2-log以3为底5lg²2+lg5·lg20log2(log3(log5(lne^125)))

来源:学生作业帮助网 编辑:作业帮 时间:2024/10/06 18:14:53
这几道对数函数题求详细解答啊……9^1/2-log以3为底5lg²2+lg5·lg20log2(log3(log5(lne^125)))
xQJ1 ɖ,AHi@AzU&$Փ*"v{$-ԛf{yFu޻]䃷I>n&Š?w<;>҂C3}0~L&܆F˯$ ~-pMvء&WH=%`dmG-K9,]@H2m-` +4.j3;؂TLW D厕%~+$gkDL+RJ鐜OvK"K3k(R dM4TV

这几道对数函数题求详细解答啊……9^1/2-log以3为底5lg²2+lg5·lg20log2(log3(log5(lne^125)))
这几道对数函数题求详细解答啊……
9^1/2-log以3为底5
lg²2+lg5·lg20
log2(log3(log5(lne^125)))

这几道对数函数题求详细解答啊……9^1/2-log以3为底5lg²2+lg5·lg20log2(log3(log5(lne^125)))
9^1/2-log以3为底5=3^[1-2log3(5)]=3^[1-log3(25)]=3/25
lg²2+lg5·lg20=(lg2)^2+lg5*(2lg2+lg5)=(lg2+lg5)^2=(lg10)^2=1
log2(log3(log5(lne^125)))=log2(log3(log5(125)))=log2(log3(3))=log2(1)=0

lg²2+lg5·lg20
= lg2 * lg2 + lg5 * (lg2 + lg2 + lg5)
= lg2 * lg2 + 2 * lg5 * lg2 + log5 * lg5
= (lg2 + lg5)^2
= [ lg(2*5) ] ^ 2
= 1