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)=
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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)=已知log(1/7)[log(3)(log(2)x
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)=
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)=
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)=
log1/7[log3(log2x)]=0=log1/7(1)
所以log3(log2x)=1
log3(log2x)=log3(3)
log2(x)=3
x=2³
x=8
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
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
关于数学对数的换底公式推论的问题已知 log(2)(3) = a,log(3(7)=b,用a,b表示log(42)(56)因为log(2)(3)=a,则1/a=log(3)(2),又∵log(3)(7)=b,∴log(42)(56)=log(3)(56)/log(3)(42)=log(3)(7)+3·log(3)(2)/log(3)(7)+log(3)(2)+1=ab+3/ab+b+1
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