Proofs of Logarithm Properties
logaM+logaN=logaMN (1)
proof:
assume: logaM=m,logaN=n
so: am=M,an=N⇒M⋅N=am⋅an=am+n
logaMN=m+n=logaM+logaN
logaM−logaN=logaMN (2)
proof:
assume: logaM=m,logaN=n
so: am=M,an=N⇒MN=aman=am−n⇒logaMN=m−n=logaM−logaN
logaaM=M (3)
proof:
assume: aM=B⇒logaB=M
so: logaaM=M
alogaM=M (4)
proof:
assume: logaM=B
so: aB=M
∵B=logaM,aB=M
∴aB=alogaM=M
logaMN=NlogaM (5)
proof:
logaMN=loga(N⏞M⋅M⋯M)
∵logaM+logaN=logaMN property (1)
∴loga(N⏞M⋅M⋯M)=N⏞logaM+logaM+⋯logaM=NlogaM
logab=logcblogca=lnblna=lgblga (6)
proof:
assume: logab=m then b=am,logcb=logcam,logcam=mlogca property (5)
∵logcb=mlogca=>logcblogca=m and m=logab
∴logab=logcblogca
logab=lnblna means c = e
logab=lgblga means c = 10
logaMbN=NMlogab (7)
proof:
∵logaMbN=logcbNlogcaM property (6)
logcbNlogcaM=NlogcbMlogca property (5)
NlogcbMlogca=NMlogcblogca
NMlogcblogca=NMlogab property (6)
∴logaMbN=NMlogab
log1ab=loga1b (8)
proof:
∵log1ab=loga−1b=loga−1b1=1−1logab property (7)
1−1logab=−1logab=−11logab=loga1b−1=loga1b
∴log1ab=loga1b
logba=1logab (9)
proof:
assume: x=logab,y=logba
⇒ax=b,by=a
x=logab=logaa⋅ba=logaa+logaba property (1)
=logaa+logaaxby=logaa+logaax−logaby property (2)
=logaa+xlogaa−ylogab property (5)
=1+x−ylogab
x=1+x−ylogab⇒ylogab=1
∵logab=x
∴yx=1⇒y=1x
∵x=logab,y=logba
∴logba=1logab
alogbc=clogba (10)
proof:
Take logb on both sides:
alogbc=clogba
⟺logbalogbc=logbclogba
⟺logbc⋅logba=logba⋅logbc property (5)
∵logbc⋅logba=logba⋅logbc
∴alogbc=clogba
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