log(a^N)(b^M)=M/Nlog(a)(b)怎么推导?

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log(a^N)(b^M)=M/Nlog(a)(b)怎么推导?

log(a^N)(b^M)=M/Nlog(a)(b)怎么推导?
log(a^N)(b^M)=M/Nlog(a)(b)怎么推导?

log(a^N)(b^M)=M/Nlog(a)(b)怎么推导?
log(a^N)(b^M)=lg(b^M)/lg(a^N)=M/N*lgb/lga=M/Nlog(a)(b)
换为同底的对数式就可以了

log(a^n)M=log(c)M /log(c)(a^n)=log(c)M /[n*log(c)a]=1/n*[log(c)M/log(c)a]=1/nlog(a)M 所以命题得证. 设左式等于数

设log(a^N)(b^M)=x,(a^N)^x=b^M,a^(Nx)=(b^M),
Nx=log(a)(b^M)=Mlog(a)(b),x=M/Nlog(a)(b)
即是log(a^N)(b^M)=M/Nlog(a)(b)

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