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Question Number 8956 by j.masanja06@gmail.com last updated on 07/Nov/16
prove that; log_(ab) x=((log_a x−log_b x)/(log_a x+log_b x))
provethat;logabx=logax−logbxlogax+logbx
Commented by sou1618 last updated on 07/Nov/16
log_p q=((log_r q)/(log_r p)) ((log_a x−log_b x)/(log_a x+log_b x))=((log_a x−(((log_a x)/(log_a b))))/(log_a x+(((log_a x)/(log_a b))))) =(( (((log_a x)/(log_a b))) )/( (((log_a x)/(log_a b))) ))×((log_a b−1)/(log_a b+1)) =((log_a b−log_a a)/(log_a b+log_a a)) =((log_a (b/a))/(log_a ab)) =log_(ab) (b/a) (=constant) ((log_a x−log_b x)/(log_a x+log_b x))≠log_(ab) x // // // // // // // // if ((log_a x−log_b x)/(log_a x+log_b x))=log_(ab) x, x=(b/a).
logpq=logrqlogrplogax−logbxlogax+logbx=logax−(logaxlogab)logax+(logaxlogab)=(logaxlogab)(logaxlogab)×logab−1logab+1=logab−logaalogab+logaa=logabalogaab=logabba(=constant)logax−logbxlogax+logbx≠logabx////////////////iflogax−logbxlogax+logbx=logabx,x=ba.

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