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Question Number 5603 by sanusihammed last updated on 22/May/16
Prove the identity log_(a/b) x = ((log_a x log_b x)/(log_b x − log_a x)) From the Right hand side ((log_a x log_b x)/(log_b x − log_a x)) Using... log_n m = ((logm)/(logn)) = ((((logx)/(loga)) × ((logx)/(logb)))/( ((logx)/(logb)) − ((logx)/(loga)))) = ((((logx)^2 )/(loga logb))/((logalogx − logb logx)/(loga logb ))) = (((logx)^2 )/(loga logb)) × ((loga logb)/(logalogx − logblogx)) = (((logx)^2 )/(logalogx − logblogx)) = (((logx)^2 )/(logx(loga − logb))) = ((logx)/(loga − logb )) = ((logx)/(log(a/b))) Using... log_n m = ((logm)/(logn)) = log_(a/b) x [Left Hand Side] PROVED THANKS SO MUCH YOZII
Provetheidentitylogabx=logaxlogbxlogbx−logaxFromtheRighthandsidelogaxlogbxlogbx−logaxUsing…lognm=logmlogn=logxloga×logxlogblogxlogb−logxloga=(logx)2logalogblogalogx−logblogxlogalogb=(logx)2logalogb×logalogblogalogx−logblogx=(logx)2logalogx−logblogx=(logx)2logx(loga−logb)=logxloga−logb=logxlogabUsing…lognm=logmlogn=logabx[LeftHandSide]PROVEDTHANKSSOMUCHYOZII
Commented by Yozzii last updated on 21/May/16
By change of base, log_n r=((log_e r)/(log_e n))=((lnr)/(ln(n))) rhs=((((lnx)/(lna))×((lnx)/(lnb)))/(((lnx)/(lnb))−((lnx)/(lna))))=((ln^2 x)/(lnalnx−lnxlnb)) rhs=((lnx)/(lna−lnb)) rhs=((lnx)/(ln(a/b)))=log_(a/b) x=lhs
Bychangeofbase,lognr=logerlogen=lnrln(n)rhs=lnxlna×lnxlnblnxlnb−lnxlna=ln2xlnalnx−lnxlnbrhs=lnxlna−lnbrhs=lnxlnab=loga/bx=lhs
Commented by sanusihammed last updated on 22/May/16
Thanks
Thanks

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