Question

高一数学，对数化简求值，求解！

求解这两题？
把解题过程发一下，谢谢

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Answer 1

lg5²+(2/3)lg8+lg5*lg30+(lg2)²
=2lg5+(2/3)lg(2³)+lg5*lg(2²*5)+(lg2)²
=2lg5+2lg2+lg5*(2lg2+lg5)+(lg2)²
=2(lg5+lg2)+2lg2*lg5+(lg5)²+(lg2)²
=2lg(5×2)+(lg5+lg2)²
=2+1
=3

9^x-6*3^x-7=0
==> (3^x)²-6*3^x-7=0
==> (3^x+1)(3^x-7)=0
==> 3^x-7=0
==> 3^x=7
==> x=log<3>7

Answer 2

(2) 解：
原式 = 2lg5 + (2/3) * 3lg2 + lg5 * ( 1+lg2) + (lg2)^2
= 2 ( 1 -lg2) + (2/3) * 3lg2 + (1- lg2) * ( 1+lg2) + (lg2)^2
= 3
(9) 解：
(3^x)^2 - 6 * 3^x - 7 = 0
( 3^x - 7 ) ( 3^x + 1 ) = 0
3^x - 7 = 0
x = log(3) 7

Answer 3

lg5^2+ (2/3)lg8 +lg5.lg20+(lg2)^2
=2lg5+ 2lg2 +(lg5)(2lg2 + lg5) +(lg2)^2
=2lg10 +2lg2.lg5 +(lg5)^2 +(lg2)^2
=2+ (lg5 + lg2)^2
=2+1
=3

Answer 4

 

Answer 5

追问

第九题是减不是加，不过还是谢谢你的回答

Answer 6

追问

非常感谢