设㏒20^625=a,求证lg32=5(4-a)/4+a

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设㏒20^625=a,求证lg32=5(4-a)/4+a
设㏒20^625=a,求证lg32=5(4-a)/4+a

设㏒20^625=a,求证lg32=5(4-a)/4+a
假设㏒a(b),a为底数,b为真数,化为自然对数,先由右边做起:a=㏒20(625)=ln625/ln20=ln5^4/ln(2*5)=(4ln5)/(2ln2+ln5) RHS=5(4-a)/(4+a) =5(4-㏒20(625))/[4+㏒20(625)] =5[4-(4ln5)/(2ln2+ln5)]/[4+(4ln5)/(2ln2+ln5)] =[20-(20ln5)/(2ln2+ln5)]/[4+(4ln5)/(2ln2+ln5)],上下通分,上下除2ln2+ln5 =[20(2ln2+ln5)-20ln5]/[4(2ln2+ln5)+4ln5],上下除4 =(10ln2+5ln5-5ln5)/(2ln2+ln5+ln5),上下除2 =5ln2/(ln2+ln5) =ln2^5/ln10 =㏒10(2^5) =㏒32=LHS