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问题：(1)利用关系式log(a)N=ba^b=N证明换底公式log(a)N=log(m)N/log(m)a(2)利用（1）中的换底公式求下式的值log(2)25*log(3)4*log(5)9(3)利用（1）中的换底公式证明log(a)b*log(b)c*log(c)a=1
答案：↓↓↓ 齐岩的回答： 网友采纳 (1)假设x=log(a)N,N=a^x假设y=log(m)N,N=m^y假设z=log(m)a,a=m^z那么N=a^x=m^y,a=m^z,代入(m^z)^x=m^y,也就是m^(zx)=m^y,即zx=y这样就有log(a)N=log(m)N/log(m)a(2)log(2)25=2log(2)5,log(3)4=2log(3)2,log(5)9=2log(5)3log(2)25*log(3)4*log(5)9=8log(2)5*log(3)2*log(5)3log(2)5=log(3)5/log(3)2原式=8log(3)5*log(5)3=8(3)log(a)b=log(c)b/log(c)/alog(a)b*log(b)c*log(c)a=log(b)c*log(c)b=log(c)c/log(c)b*log(c)b=log(c)c=1a | 
