f(log2 1/3)=-(2^(-log2 1/3)-1)=-2 怎么算

来源：学生作业帮编辑：作业帮分类：数学作业时间：2024/07/16 14:08:14

f(log2 1/3)=-(2^(-log2 1/3)-1)=-2 怎么算
就是 -(2^(-log2 1/3)-1)=-2 是怎么来的

设a^(loga b)=x,式子两边取底数为a的对数 ,
即loga(a^(loga b))=loga x
loga b=loga x
得x=b
所以
-(2^(-log2 1/3)-1)
=-(2^(-log2 3^-1)-1)
=-(2^(log2 3)-1)
=-(3-1)
=-2

log2 (x + 3) + log2(x + 2) = 1 (log2 9)/(log2 8) = 2/3(log2 3) 已知函数f(x)=log2(2-x)+log2(2+x),g(x)=log2(2x-1) 2^(log2^(-3))= 解方程log2(2-x)=log2(x-1)+1 2^log2 (3) × log2 (1/8) + 2lg3 不等式log2(2x+3)>log2(x+1)的解集是求函数f(x)=(log2 4x)(log2 2x)在1/4 log2(3)*log3(4)*log4(5)*.log(k+1)(k+2)=log2(k+2) 怎么证 log2(1+根号2+根号3）+log2（1+根号2-根号3）= 方程log2[9^(x-2) +7]=2+log2[3^(x-1) +1]的解为 log2(2^x-1)log2(2^(x+2)-4=3