若3^x=2^y且1/x+1/y=2,則3^x+2^y=?
回答 (2)
2017-11-27 1:54 pm
x log 3 = y log 2
y= x log 3 / log 2
1/x+ 1/y= (1/x) (1+ log 2 /log 3) = 2
x = (1+ log 2 /log 3) / 2
y = (log 3 /log 2)(1+ log 2 /log 3) / 2
3^x = 3 ^[ (1+ log 2 /log 3) / 2] = 3^ [(log 3 + log 2)/(2 log 3)] = 10^ [(log 3 + log 2)/2] = √6 = 2^y
3^x+2^y= = 2 √6 ....Ans
y= x log 3 / log 2
1/x+ 1/y= (1/x) (1+ log 2 /log 3) = 2
x = (1+ log 2 /log 3) / 2
y = (log 3 /log 2)(1+ log 2 /log 3) / 2
3^x = 3 ^[ (1+ log 2 /log 3) / 2] = 3^ [(log 3 + log 2)/(2 log 3)] = 10^ [(log 3 + log 2)/2] = √6 = 2^y
3^x+2^y= = 2 √6 ....Ans
2017-11-27 6:19 am
Sol
設3^x=2^y=p>0
xlog3=ylog2=logp
x=logp/log3
1/x=log3/logp
1/y=log2/logp
1/x+1/y=2
log3/logp+log2/logp=2
log6/logp=2
log6=2logp
log6=logp^2
p^2=6
3^x+2^y
=2p
=2√6
設3^x=2^y=p>0
xlog3=ylog2=logp
x=logp/log3
1/x=log3/logp
1/y=log2/logp
1/x+1/y=2
log3/logp+log2/logp=2
log6/logp=2
log6=2logp
log6=logp^2
p^2=6
3^x+2^y
=2p
=2√6
收錄日期: 2021-04-30 22:33:49
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20171126114220AAeDMtM