已知log2[log1/2(log2x)]=log3[log1/3(log3y)]=log5[log1/5(log5z)]=0.求x2+y^3-z^5的值
1个回答
展开全部
log2[log1/2(log2x)]=0=log2(1)
log1/2(log2x)=1=log1/2(1/2)
log2(x)=1/2
x=2^(1/2)
log3[log1/3（log3y）]=0
log1/3（log3y）=1
log3(y)=1/3
y=3^(1/3)
x^6=2^3=8
y^6=3^2=9
所以y>x
log5[log1/5(log5z)]=o
log1/5(log5z)=1
log5(z)=1/5
z=5^(1/5)
x^2=2
y^3=3
z^5=5
x2+y^3-z^5=2+3-5=0
log1/2(log2x)=1=log1/2(1/2)
log2(x)=1/2
x=2^(1/2)
log3[log1/3（log3y）]=0
log1/3（log3y）=1
log3(y)=1/3
y=3^(1/3)
x^6=2^3=8
y^6=3^2=9
所以y>x
log5[log1/5(log5z)]=o
log1/5(log5z)=1
log5(z)=1/5
z=5^(1/5)
x^2=2
y^3=3
z^5=5
x2+y^3-z^5=2+3-5=0
本回答被提问者采纳
已赞过
已踩过<
评论
收起
你对这个回答的评价是？
推荐律师服务：
若未解决您的问题，请您详细描述您的问题，通过百度律临进行免费专业咨询
