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Question Number 5988 by sanusihammed last updated on 08/Jun/16

$$Ifur=logr$$$$showthat\sum _{r=1}^{10}ur=log3628800$$$$$$$$pleasehelp.$$

Answered by Yozzii last updated on 08/Jun/16

$$ur=logr$$$$\therefore \underset{r=1}{\sum ^{10}}ur=\underset{r=1}{\sum ^{10}}logr=S$$$$Forx,y>0wehavethatlogx+logy=logxy.$$$$$$$$PROOF:$$$$Letu=a^{{log}_{a}m}wherem>0\mathrm{\&}a>0.\Rightarrow u>0$$$$Alsoletn={log}_a{r}^k(a>0,r>0,k\ne 0)\Rightarrow r^k=a^n\Rightarrow r=a^{n/k}\Rightarrow {log}_{a}r=n/k\Rightarrow n={klog}_{a}r\left(Powerrule\right)$$$$\Rightarrow {log}_{a}u={log}_a\left(a^{{log}_{a}m}\right)=\left({log}_{a}m\right)\left({log}_{a}a\right)$$$$Sincea=a^1\Rightarrow {log}_{a}a=1.$$$${log}_{a}u={log}_{a}m\Rightarrow u=m\therefore m=a^{{log}_{a}m}.$$$$So,ifx,y>0\Rightarrow a^{{log}_{a}x}\times a^{{log}_{a}y}=xyor{a}^{{log}_{a}x+{log}_{a}y}=xy$$$$\Rightarrow {log}_{a}xy={log}_a{a}^{{log}_{a}x+{log}_{a}y}=({log}_{a}x+{log}_{a}y)\left({log}_{a}a\right)$$$$\Rightarrow {log}_{a}xy={log}_{a}x+{log}_{a}y◻$$$$$$$$Thus,S=log\underset{r=1}{\prod ^{10}}r=log(10!)=log3628800$$

Commented by sanusihammed last updated on 08/Jun/16

$$Wowthanks.ireallyappreiate=$$

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