高中数学,对数问题,求详细过程及答案,必采纳,先谢过了,先到先得~
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/ , 斜线左边是分子,右边是分母
1) 3^log3(√5)+√3^log3(1/5)
=3^log3(√5)+3^[1/2log3(1/5)]
=3^log3(√5)+3^[log3(√1/5)]
=√5+(√1/5)
=√5+(√5/5)
=6√5/5
2)(lg2+lg5-lg8)/(lg50-lg90)
=lg[2*5)/8]/lg(50/90)
=lg(5/4)/lg(5/9)
=[lg5/lg4]*[[lg9/lg5]
=lg9/lg4
=lg3/lg2
=log2(3)
3)log3(4√27/3)*log5[4^1/2log2(10)-(3√3)^(2/3)-7^log7(2)]
=)log3(4*3√3/3)*log5[2^(2*1/2)log2(10)-3^(3/2*2/3)-7^log7(2)]
=log3(4√3)*log5[2^log2(10)-3-2]
=log3(4√3)*log5(10-3-2)
=log3(4√3)*log5(5)
=log3(4√3)*1
=log3(4)+log3(√3)
=log3(4)+(1/3)
1) 3^log3(√5)+√3^log3(1/5)
=3^log3(√5)+3^[1/2log3(1/5)]
=3^log3(√5)+3^[log3(√1/5)]
=√5+(√1/5)
=√5+(√5/5)
=6√5/5
2)(lg2+lg5-lg8)/(lg50-lg90)
=lg[2*5)/8]/lg(50/90)
=lg(5/4)/lg(5/9)
=[lg5/lg4]*[[lg9/lg5]
=lg9/lg4
=lg3/lg2
=log2(3)
3)log3(4√27/3)*log5[4^1/2log2(10)-(3√3)^(2/3)-7^log7(2)]
=)log3(4*3√3/3)*log5[2^(2*1/2)log2(10)-3^(3/2*2/3)-7^log7(2)]
=log3(4√3)*log5[2^log2(10)-3-2]
=log3(4√3)*log5(10-3-2)
=log3(4√3)*log5(5)
=log3(4√3)*1
=log3(4)+log3(√3)
=log3(4)+(1/3)
追问
那个,老师把题给错了,纠正之后我自己会做了
谢谢'你啊
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题目抄的看不懂,那个lg后。。。。
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追答
你直接把资料上拍下来吧
早上6点给你做
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