设x,y,z属于R,且3^x=4^y=6^z【要求详细过程】

(1)证明：
3^x = 4^y = 6^z
3^x = 2²^y = (2^z)×(3^z)
∴(3^x)×(2^2y) = (2^z)×(3^z)×(2^z)×(3^z)
(3^x)×(2^2y) = (2^2z)×(3^2z)
∵ 3^n 为奇数,2^n为偶数
∴ 3^x=3^2z,2^2y=2^2z,
⅞x=2z,2y=2z
∴1/(2y)=1/(2z),1/x=1/(2z)
于是1/z - 1/x=1/z-1/(2z)=1/(2z)=1/(2y)
即1/z - 1/x=1/（2y)



设3^x=4^y=6^z=k,则x=log3,k,y=log4,k,z=log6,k
很显然k>1,从而
x=1/logk,3 y=1/logk,4 z=1/logk,6
所以3x=3/log(k)3,4y=4/log(k)4,6z=6/log(k)6
3x/4y=[3logk,4]/[4logk,3]=logk,4³/logk,3⁴=logk,64/logk,81<1
所以3x<4y
同理有
4y/6z=[4logk,6]/[6logk,4]=logk,1296/logk,4096<1
所以4y<6z
所以3x<4y<6z

设3^x=4^y=6^z=M
log3(m)=x 1/x=logm(3)
log4(m)=y 1/y=logm(4)
log6(m)=z 1/z=logm(6)
1/z-1/x=logm(6)-logm(3)=logm(2)=1/2*logm(4)=1/2y
1/(3x)=1/3*logm(3)
1/(4y)=1/4*logm(4)
1/(6z)=1/6*logm(6)
又3次根号下3>4次根号下4>6次根号下6
所以3x<4y<6z

取对数求解