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加入日期: Apr 2004
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土法煉鋼
∵2^x=3^y=6^z
∴log2^x=log3^y=log6^z (1)
∵loga^b=bloga
由(1),令xlog2=ylog3=zlog6=k
yz+zx-xy
=(k/log3)(k/log6)+(k/log6)(k/log2)-(k/log2)(k/log3)
={(1/log6)[(1/log3)+(1/log2)]-(1/log2)(1/log3)}k^2
={(1/log6)[(log2+log3)/(log3log2)]-(1/log2)(1/log3)}k^2
={(1/log6)[log6/(log3log2)]-(1/log2)(1/log3)}k^2
={[1/(log3log2)]-(1/log2)(1/log3)}k^2
=0
     
      
舊 2006-11-15, 11:16 AM #11
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