已知abc均为正数学且满足3^a=4^b=6^c则A.1/c=1/a+1/bB.1/c=2/a+2/bC.2/c=1/a+2/bD.2/c=2/a+1/b已知abc均为正数学且满足3^a=4^b=6^c则A.1/c=1/a+1/bB.1/c=2/a+2/bC.2/c=1/a+2/bD.2/c=2/a+1/b

Question

已知a b c 均为正数学且满足3^a=4^b=6^c 则 A.1/c=1/a+1/b B.1/c=2/a+2/b C.2/c=1/a+2/b D.2/c=2/a+1/b
已知a b c 均为正数学且满足3^a=4^b=6^c 则
A.1/c=1/a+1/b B.1/c=2/a+2/b C.2/c=1/a+2/b D.2/c=2/a+1/b

Answer 1

选D.
令3^a=4^b=6^c=n
则a=log3(n) 1/a=logn(3)
同理,1/b=logn(4) 1/c=logn(6)
1/b=2logn(2) 1/(2b)=logn(2)
由于logn(2)+logn(3)=logn(6)
因此1/(2b)+1/a=1/c
等式两边同乘以2
1/b+2/a=2/c
选D.