Question

设2^a=5^b=m,且1/a+1/b=2,则m=？

Answer 1

因为logab*logba=1，所以logab=1/logba
所以2^a=5^b=m
a=log2m，b=log5m
1/a=logm2 1/b=logm5
所以1/a+1/b=logm2+logm5=logm10=2
所以m=√10

Answer 2

因为2^a=5^b=m,
所以log2^a=log5^b=logm,
所以alog2=blog5=logm
因为1/a+1/b=2
所以b=a/(2a-1)
所以alog2=blog5
alog2=alog5/(2a-1)
2a-1=log5/log2
a=(log(5)2+1)/2
所以logm=alog2=(log(5)2+1)log2/2
m=e^((log(5)2+1)log2/2)

Answer 3

设2^a=5^b=m，且 1/a+1/b=2，则m=（ ）为什么 1/a+1/b =所以logab=1/logba 所以2^a=5^b=m a=log2m，b=log5m 1/a=logm2 1/

Answer 4

2^a=5^b=m,

lg2^a=lg5^b=lgm
alg2=blg5=lgm
a=lgm/lg2
b=lgm/lg5

1/a+1/b=2
1/(lgm/lg2)+1/(lgm/lg5)=2
lg2/lgm+lg5/lgm=2
(lg2+lg5)/lgm=2
1/lgm=2
lgm=1/2
m=10^(1/2)=√10