Hei, noen som kan hjelpe meg med en oppgave litt fort?
Oppgaven er at jeg skal vise at a) n^2* (x/n)= x^2, blir b) lg(x/n)^lgx=lg(x/n)^2. Dette har jeg fått til, men så er det videre å vise at b) blir c) (lgx-n)*(lgx-lgn)=0. Jeg får ikke til den utregningen fra b til c
Logaritme oppgave (trenger hjelp fort)
Moderatorer: Aleks855, Gustav, Nebuchadnezzar, Janhaa, DennisChristensen, Emilga
Antar at du mener [tex]n^n*\left ( \frac{x}{n} \right )^{lgx}=x^n[/tex]
[tex]\frac{n^n}{n^n}*\left ( \frac{x}{n} \right )^{lgx}=\frac{x^n}{x^n}[/tex]
[tex]\left ( \frac{x}{n} \right )^{lgx}=\frac{x^n}{n^n}[/tex]
[tex]\left ( \frac{x}{n} \right )^{lgx}=\left (\frac{x}{n} \right )^n[/tex]
[tex]lgx \,\,lg\left ( \frac{x}{n} \right )=n\,\, lg\left (\frac{x}{n} \right )[/tex]
[tex]lgx \,\,\left (lgx-lgn \right )=n\,\, (lgx-lgn)[/tex]
[tex]lgx \,\,\left (lgx-lgn \right )=n\,\, (lgx-lgn)\Longleftrightarrow lgx{\color{Red} {(lgx-lgn)}}-n{\color{Red} {(lgx-lgn)}}=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\color{Red} {(lgx-lgn)}}(lgx-n)=0[/tex]
[tex]\frac{n^n}{n^n}*\left ( \frac{x}{n} \right )^{lgx}=\frac{x^n}{x^n}[/tex]
[tex]\left ( \frac{x}{n} \right )^{lgx}=\frac{x^n}{n^n}[/tex]
[tex]\left ( \frac{x}{n} \right )^{lgx}=\left (\frac{x}{n} \right )^n[/tex]
[tex]lgx \,\,lg\left ( \frac{x}{n} \right )=n\,\, lg\left (\frac{x}{n} \right )[/tex]
[tex]lgx \,\,\left (lgx-lgn \right )=n\,\, (lgx-lgn)[/tex]
[tex]lgx \,\,\left (lgx-lgn \right )=n\,\, (lgx-lgn)\Longleftrightarrow lgx{\color{Red} {(lgx-lgn)}}-n{\color{Red} {(lgx-lgn)}}=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\color{Red} {(lgx-lgn)}}(lgx-n)=0[/tex]
[tex]i*i=-1[/tex]
Omnia mirari etiam tritissima - Carl von Linné
( Find wonder in all things, even the most commonplace.)
Det er åpning og lukking av ionekanaler i nerveceller som gjør det mulig for deg å lese dette.
Omnia mirari etiam tritissima - Carl von Linné
( Find wonder in all things, even the most commonplace.)
Det er åpning og lukking av ionekanaler i nerveceller som gjør det mulig for deg å lese dette.