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Question-10898




Question Number 10898 by Saham last updated on 01/Mar/17
Answered by sandy_suhendra last updated on 02/Mar/17
E^2 =m^2 c^b +(pc)^n   m^2 c^b =E^2 −(pc)^n   c^b =((E^2 −(pc)^n )/m^2 )  ln c^b =ln [((E^2 −(pc)^n )/m^2 )]  b ln c = ln [E^2 −(pc)^n ]−2 ln m  b=((ln [E^2 −(pc)^n ]−2 ln m)/(ln c))  or   b =^c  log [((E^2 −(pc)^n )/m^2 )]    E^2 =m^2 c^b +(pc)^n   (pc)^n =E^2 −m^2 c^b   ln (pc)^n =ln (E^2 −m^2 c^b )  n ln pc = ln (E^2 −m^2 c^b )  n=((ln (E^2 −m^2 c^b ))/(ln pc))  n=((ln (E^2 −m^2 c^b ))/(ln p + ln c))  or   n =^(pc)  log (E^2 −m^2 c^b )
$$\mathrm{E}^{\mathrm{2}} =\mathrm{m}^{\mathrm{2}} \mathrm{c}^{\mathrm{b}} +\left(\mathrm{pc}\right)^{\mathrm{n}} \\ $$$$\mathrm{m}^{\mathrm{2}} \mathrm{c}^{\mathrm{b}} =\mathrm{E}^{\mathrm{2}} −\left(\mathrm{pc}\right)^{\mathrm{n}} \\ $$$$\mathrm{c}^{\mathrm{b}} =\frac{\mathrm{E}^{\mathrm{2}} −\left(\mathrm{pc}\right)^{\mathrm{n}} }{\mathrm{m}^{\mathrm{2}} } \\ $$$$\mathrm{ln}\:\mathrm{c}^{\mathrm{b}} =\mathrm{ln}\:\left[\frac{\mathrm{E}^{\mathrm{2}} −\left(\mathrm{pc}\right)^{\mathrm{n}} }{\mathrm{m}^{\mathrm{2}} }\right] \\ $$$$\mathrm{b}\:\mathrm{ln}\:\mathrm{c}\:=\:\mathrm{ln}\:\left[\mathrm{E}^{\mathrm{2}} −\left(\mathrm{pc}\right)^{\mathrm{n}} \right]−\mathrm{2}\:\mathrm{ln}\:\mathrm{m} \\ $$$$\mathrm{b}=\frac{\mathrm{ln}\:\left[\mathrm{E}^{\mathrm{2}} −\left(\mathrm{pc}\right)^{\mathrm{n}} \right]−\mathrm{2}\:\mathrm{ln}\:\mathrm{m}}{\mathrm{ln}\:\mathrm{c}} \\ $$$$\mathrm{or}\:\:\:\mathrm{b}\:=^{\mathrm{c}} \:\mathrm{log}\:\left[\frac{\mathrm{E}^{\mathrm{2}} −\left(\mathrm{pc}\right)^{\mathrm{n}} }{\mathrm{m}^{\mathrm{2}} }\right] \\ $$$$ \\ $$$$\mathrm{E}^{\mathrm{2}} =\mathrm{m}^{\mathrm{2}} \mathrm{c}^{\mathrm{b}} +\left(\mathrm{pc}\right)^{\mathrm{n}} \\ $$$$\left(\mathrm{pc}\right)^{\mathrm{n}} =\mathrm{E}^{\mathrm{2}} −\mathrm{m}^{\mathrm{2}} \mathrm{c}^{\mathrm{b}} \\ $$$$\mathrm{ln}\:\left(\mathrm{pc}\right)^{\mathrm{n}} =\mathrm{ln}\:\left(\mathrm{E}^{\mathrm{2}} −\mathrm{m}^{\mathrm{2}} \mathrm{c}^{\mathrm{b}} \right) \\ $$$$\mathrm{n}\:\mathrm{ln}\:\mathrm{pc}\:=\:\mathrm{ln}\:\left(\mathrm{E}^{\mathrm{2}} −\mathrm{m}^{\mathrm{2}} \mathrm{c}^{\mathrm{b}} \right) \\ $$$$\mathrm{n}=\frac{\mathrm{ln}\:\left(\mathrm{E}^{\mathrm{2}} −\mathrm{m}^{\mathrm{2}} \mathrm{c}^{\mathrm{b}} \right)}{\mathrm{ln}\:\mathrm{pc}} \\ $$$$\mathrm{n}=\frac{\mathrm{ln}\:\left(\mathrm{E}^{\mathrm{2}} −\mathrm{m}^{\mathrm{2}} \mathrm{c}^{\mathrm{b}} \right)}{\mathrm{ln}\:\mathrm{p}\:+\:\mathrm{ln}\:\mathrm{c}} \\ $$$$\mathrm{or}\:\:\:\mathrm{n}\:=^{\mathrm{pc}} \:\mathrm{log}\:\left(\mathrm{E}^{\mathrm{2}} −\mathrm{m}^{\mathrm{2}} \mathrm{c}^{\mathrm{b}} \right) \\ $$
Commented by Saham last updated on 02/Mar/17
God bless you sir.
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

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