已知log(1/7)[log(3)(log(2)x)]=0已知log1/7[log3(log2x)]=0(1/7;3;2都为底数),x^(-1/2)=
问题描述：
已知log(1/7)[log(3)(log(2)x)]=0
已知log1/7[log3(log2x)]=0(1/7;3;2都为底数),x^(-1/2)=
答
log1/7[log3(log2x)]=0
则Iog1/7(1)=o
所以Iog3(Iog2x)=1（Iog2x=3 Iog3(3）=1
固x=8
x^(-1/2)=-2根号2
答
先解方程，一层一层从外向里解
最外层log(1/7)[log(3)(log(2)x)]=0
可以得到log(3)(log(2)x)=(1/7)^0=1
然后进而得到log(2)x=3^1=3
最后有x=2^3=8
所以x^(-1/2)=8^(-1/2)=1/(2√2)=(√2)/4
答
log1/7[log3(log2x)]=0=log1/7(1)
所以log3(log2x)=1
log3(log2x)=log3(3)
log2(x)=3
x=2³
x=8