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2^(1+log2^3)=?

=2^1 *2^(log2^3) =2*3=6

log1/2(2/3) =log1/2(2)-log1/2(3) =-1+lg2(3) =-1+1.585 =0.585

解: log2(2x-3)

如图所示

1、解:由,log2^x+log2^(3*2^n-1-x)大于等于2n-1 去对数得: x*[3*2^(n-1)-x]》2^(2n-1) 即:2x²-(3*2^n)*x+2^(2n)《0 (2x-2^n)*(x-2^n)《0 所以, 2^(n-1)《x《2^n 再由题定义域: 0

f(x)=log (x^2+a) f(3)=1 log (9+a) =1 9+a =2 a=-7

可化为(2^x-3^(x+1))(2^x+3^x)=0 =>2^x=3*3^x =>(2/3)^(x)=3 =>x=log(2/3)[3]

解:已知x+1/x=3,则(x+1/x)^2=9,即: x^2+x^-2=9-2=7 x-1/x=√(x-1/x)^2=√(x^2-2*x*1/x+x^-2)=√5 x^2-x^-2=(x+1/x)(x-1/x)=3*√5=3√5 log(4)9-log(5)4xlog(8)5 =log(2^2)3^2-log(5)2^2*log(2^3)5 =log(2)3-2/3*log(5)2*log(2)5 =log(2)3-2/3

(1)由2log3^x+log3²=(1/2)log3^16 log3^2x²=log3^4 ∴2x²=4 x=√2(x=-√2舍去) (2)由log2^x≤-1/2 ∵x≤2^-½ 0<x≤√2/2, (3)由log2^x+log2^(x-8)=7 log2^(x(x-8)=7 ∴x(x-8)=2^7 x²-8x-128=0 (x-16)(x+8)=0 x=16,(x=-8舍去...

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