Jule-nøtt liknings-system

[Janhaa]

Er såpass hyper nå, at jeg slenger inn en morsom oppgave, vdr likningssystem;

Sjekk fila/bildet.

Arbeidet skal jo vises

D3BAA0B2-6BF9-4B8F-B1BE-AABCA2DCD2A1.jpeg (41.54 kiB) Vist 41424 ganger

[Kay]

[tex]x^y=4^6\Rightarrow \sqrt[y]{x^y}=\sqrt[y]{4^6}\Rightarrow x=4^{\frac{6}{y}}[/tex][tex]y=1+\log_4x=1+\log_4 4^{\frac{6}{y}}=1+\frac{6}{y}\Rightarrow y^2-y-6=0\Rightarrow y=3 \ \ \vee \ \ y=-2[/tex]

Så at [tex]x^3=4^{6}\Rightarrow x=16[/tex] som eneste reelle løsning og [tex]x^{-2}=4^6\Rightarrow x=\frac{1}{64}[/tex] som eneste reelle positive løsning.

Så vi har løsningene [tex](x_1,y_1)=(16,3)[/tex] og [tex](x_2,y_2)=\left(\frac{1}{64},-2\right)[/tex]

[Mattebruker]

Alternativ løysing:

I innsett i II gir

x [tex]^{( 1 + log_{4}( x ))}[/tex] = 4[tex]^{6}[/tex] ( x = 4[tex]^{log_{4}( x )}[/tex] ) [tex]\Leftrightarrow[/tex] 4[tex]^{log_{4}( x )( 1+ log_{4}(x)) }[/tex] = 4[tex]^{6}[/tex]

[tex]\Leftrightarrow[/tex] ( Sett log[tex]_{4}[/tex]( x ) = z ) z[tex]^{2}[/tex] + z = 6 [tex]\Leftrightarrow[/tex] z = -3 v z = 2 [tex]\Leftrightarrow[/tex] x = 4[tex]^{-3}[/tex] = [tex]\frac{1}{64}[/tex] v x = 4[tex]^{2}[/tex] = 16

[Janhaa]

Fine løsninger

[SpreVitenskapVidere]

\begin{align*}
& \begin{cases}
&(1)\quad y =1+\log_{4} x\\
&(2) x^{y}=4^6\\
\end{cases}\\
&(2) \log _{4}x^{y}=\log_{4}(4^6)\\
&y\cdot \log_{4}x=6\cdot log_{4} 4\\
& y \log_{4}x=6\cdot 1\\
& y=\frac{6}{\log _{4}(x)}\\
&(1) \dfrac{6}{\log _{4}(x)}=1+\log_{4} x\quad \vert \cdot \log _{4}(x) \\
&6= \log_{4}x+\left( \log_{4}x\right) ^{2}\\
&\left( \log_{4}x\right) ^{2}+ \log_{4}x-6=0\\
&La\quad u= \log_{4}x\\
& u^{2}+u-6=0\\
&sum: +1\\
&gang: -6\\
&\left( u+3\right) \left( u-2\right) =0\\
&u+3=0\Rightarrow u=-3\Rightarrow \log_{4}x=-3\\
& x=4^{-3}=\dfrac{1}{4^{3}}=\dfrac{1}{64}\\
& y=\frac{6}{\log _{4}(4^-3)}=\frac{6}{-3\cdot\log _{4}(4)}=\frac{6}{-3\cdot1}=-2\\
& eller\\
&u-2=0\Rightarrow u=2\Rightarrow \log_{4}x=2\\
& x=4^{2}=16\\
& y=\frac{6}{\log _{4}(4^2)}=\frac{6}{2\cdot\log _{4} 4}=\frac{6}{2\cdot1}=3\\
\end{align*}
Løsninger blir
$$\{ (x=\frac{1}{64},y=-2),(x=16,y=3)\}$$

[Janhaa]

\begin{align*}
& \begin{cases}
&(1)\quad y =1+\log_{4} x\\
&(2) x^{y}=4^6\\
\end{cases}\\
&(2) \log _{4}x^{y}=\log_{4}(4^6)\\
&y\cdot \log_{4}x=6\cdot log_{4} 4\\
& y \log_{4}x=6\cdot 1\\
& y=\frac{6}{\log _{4}(x)}\\
&(1) \dfrac{6}{\log _{4}(x)}=1+\log_{4} x\quad \vert \cdot \log _{4}(x) \\
&6= \log_{4}x+\left( \log_{4}x\right) ^{2}\\
&\left( \log_{4}x\right) ^{2}+ \log_{4}x-6=0\\
&La\quad u= \log_{4}x\\
& u^{2}+u-6=c\\
&sum: +1\\
&gang: -6\\
&\left( u+3\right) \left( u-2\right) =0\\
&u+3=0\Rightarrow u=-3\Rightarrow \log_{4}x=-3\\
& x=4^{-3}=\dfrac{1}{4^{3}}=\dfrac{1}{64}\\
& y=\frac{6}{\log _{4}(4^-3)}=\frac{6}{-3\cdot\log _{4}(4)}=\frac{6}{-3\cdot1}=-2\\
& eller\\
&u-2=0\Rightarrow u=2\Rightarrow \log_{4}x=2\\
& x=4^{2}=16\\
& eller\\
& y=\frac{6}{\log _{4}(4^2)}=\frac{6}{2\cdot\log _{4} 4}=\frac{6}{2\cdot1}=3\\
\end{align*}
Løsninger blir
$$\{ (x=\frac{1}{64},y=-2),(x=16,y=3)\}$$

Takk for bidraget ditt… var den omstendelig løsninga jg sjøl har!
Hadde tenkt å legge den inn, men der kom den ;=)