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If-ur-log-r-show-that-r-1-10-ur-log3628800-please-help-

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Question Number 5988 by sanusihammed last updated on 08/Jun/16
If ur = log r show that Σ_(r = 1) ^(10) ur = log3628800 please help.
Ifur=logrshowthat∑r=110ur=log3628800pleasehelp.
Answered by Yozzii last updated on 08/Jun/16
ur=logr ∴Σ_(r=1) ^(10) ur=Σ_(r=1) ^(10) logr=S For x,y>0 we have that logx+logy=logxy. PROOF: Let u=a^(log_a m) where m>0 & a>0.⇒u>0 Also let n=log_a r^k (a>0,r>0,k≠0)⇒r^k =a^n ⇒r=a^(n/k) ⇒log_a r=n/k⇒n=klog_a r (Power rule) ⇒log_a u=log_a (a^(log_a m) )=(log_a m)(log_a a) Since a=a^1 ⇒log_a a=1. log_a u=log_a m⇒u=m ∴ m=a^(log_a m) . So, if x,y>0⇒ a^(log_a x) ×a^(log_a y) =xy or a^(log_a x+log_a y) =xy ⇒log_a xy=log_a a^(log_a x+log_a y) =(log_a x+log_a y)(log_a a) ⇒log_a xy=log_a x+log_a y □ Thus, S=logΠ_(r=1) ^(10) r=log(10!)=log3628800
ur=logr∴∑10r=1ur=∑10r=1logr=SForx,y>0wehavethatlogx+logy=logxy.PROOF:Letu=alogamwherem>0&a>0.⇒u>0Alsoletn=logark(a>0,r>0,k≠0)⇒rk=an⇒r=an/k⇒logar=n/k⇒n=klogar(Powerrule)⇒logau=loga(alogam)=(logam)(logaa)Sincea=a1⇒logaa=1.logau=logam⇒u=m∴m=alogam.So,ifx,y>0⇒alogax×alogay=xyoralogax+logay=xy⇒logaxy=logaalogax+logay=(logax+logay)(logaa)⇒logaxy=logax+logay◻Thus,S=log∏10r=1r=log(10!)=log3628800
Commented by sanusihammed last updated on 08/Jun/16
Wow thanks. i really appreiate=
Wowthanks.ireallyappreiate=

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