已知log2[3]=a,log3[7]=b,用a,b表示log42[17]

没有错...换底公式的运用于逆运用.log(2)(3)xlog(3)(7)=ln3xln7/ln2xln3=ln7/ln2=log(2)(7)

题目错了

log23=a;log37=b即lg3/lg2=a;lg7/lg3=b;所以lg7/lg2=a*b所以log72=1/a*b同理log142=1/1+ab;所以结果为（3+ab）/(1+ab)

运用换底:log14(56)=log3(56)/log3(14)=〔log3(7)＋log3(8)〕/〔log3(7)＋log3(2)〕log3(2)=1/log2(3)=1/alog3(8)=3log3(2)=3/a原式=(b＋3/a)/

log14^56=log3^56/log3^14=(log3^7+3log3^2)/(log3^7+log3^2)=(b+3/a)/(b+1/a)=(3+ab)/(1+a)

运用换底:log14(56)=log3(56)/log3(14)=〔log3(7)＋log3(8)〕/〔log3(7)＋log3(2)〕log3(2)=1/log2(3)=1/alog3(8)=3log3(2)=3/a原式=(b＋3/a)/

log1256=lg56/lg12=(3lg2+lg7)/(lg3+lg4)因为a=lg3/lg2所以lg3=alg2因为b=lg7/lg3所以lg7=blg3=ablg2所以原式=(3lg2+ablg2)/(alg2+2lg2)=ab+3

a=lg3/lg2b=lg7/lg3log4256=lg56/lg42=(lg7+3lg2)/(lg2+lg3+lg7)=(lg7/lg3+3lg2/lg3)/(lg2/lg3+1+lg7/lg3)=(b+3/a)/(1/a+1+b)=(a

log2(3)=a,则log3(2)=1/a.则log36(45)=[log3(36)]/[log3(45)]=[log3(4)＋log3(9)]/[log3(9)＋log3(5)]=[2log3(2)＋2]/[2＋log3(5)]=[2＋

运用换底log14(56)=log3(56)/log3(14)=log3(7*8)/log3(2*7)=〔log3(7)＋log3(8)〕/〔log3(7)＋log3(2)〕log3(2)=1/log2(3)=1/alog3(8)=3log

log23=lg3/lg2=a,所以lg3=alg2,同理：log37=lg7/lg3=b,所以lg7=blg3即lg7=ab*lg2log1456=lg56/lg14其中lg56=lg7+lg8=lg7+3lg2=lg2(ab+3)lg1

log2(3)+log3(5)+log3(2)=log2(3)+log3(10)=4.2429387700296再问：=1g3/1g2+(lg5+lg2)lg3这个式子最后那个lg3原来是分母，，怎么变成分子了=lg3(1g2+1g10)还

倒数第二个式子分子分母同时提出来一个lg2然后约掉了啊

等于2/a利用换底公式,log3^2=log2^2/log2^3=a,即log2^3=1/a;log2^9=log2^(3^2)=2log2^3=2*(1/a)=2/a

log(2)3=a==>lg3/lg2=a==>lg3=alg2log(3)5=b==>lg5/lg3=b==>lg5=blg3=ablg2log(15)20=lg20/lg15=(lg2+lg2+lg5)/(lg3+lg5)=(lg2+l

a=lg3/lg2,lg2=lg3/ab=lg5/lg3,lg5=blg3log15(20)=lg20/lg15=lg(2²*5)/lg(3*5)=(2lg2+lg5)/(lg3+lg5)=把前面带入,约去lg3=(ab+2)/(

log₂3=a,==>log₃2=1/alog₃7=b用换底公式：log3√7^（2√21）=log₃(2√21)/log₃(3√7)=[log₃2+1/2(log&#

已知log210=a,换底公式1/lg2=alg2=1/alog310=b1/lg3=blg3=1/blog3(4)=lg4/lg3=2lg2/lg3=2b/a请参考……

log3(2)=lg2/lg3=a/blog3(4)=2log3(2)=2a/

log23=mlg3/lg2=mlg3=mlg2同理lg7=nlg3所以lg7=mnlg2log4256=(lg56)/(lg42)=(lg8*7)/(lg2*3*7)=(3lg2+lg7)/(lg2+lg3+lg7)=(3lg2+mnlg