Question

已知log2 3=a,log3 7=b,试用a,b表示log14 56

Answer 1

解：运用换底:log14(56)=log3(56)/log3(14)
=〔log3(7)＋log3(8)〕/〔log3(7)＋log3(2)〕

log3(2)=1/log2(3)=1/a
log3(8)=3log3(2)=3/a

原式=(b＋3/a)/(b＋1/a)
=(ab＋3)/(ab＋1)
=1＋2/(ab＋1)

Answer 2

换底:log14(56)=log3(56)/log3(14)
=〔log3(7)＋log3(8)〕/〔log3(7)＋log3(2)〕

log3(2)=1/log2(3)=1/a
log3(8)=3log3(2)=3/a

原式=(b＋3/a)/(b＋1/a)
=(ab＋3)/(ab＋1)
=1＋2/(ab＋1)

Answer 3

log2
3=a;log
3
7=b
即lg3/lg2=a;lg7/lg3=b;
所以lg7/lg2=a*b
所以log7
2=1/a*b同理log14
2=1/1+ab;
所以结果为（3+ab）/(1+ab)

Answer 4

若log2
3=a，log3
7=b
则log2
3*log3
7=ab
即log2
7=ab
log14
56
=log2
14/(log2
56)
=(log2
2*7)/(log2
7*8)
=(log2
2+log2
7)/(log2
7+log2
8)
=(1+log2
7)/(log2
7+3)
=(ab+1)/(ab+3)

Answer 5

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