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Daniel za post:
Nevena12
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od Daniel » Ponedeljak, 01. Jul 2013, 20:39
Prvo postavimo uslov za definisanost: [inlmath]x>0[/inlmath]
Logaritmujemo za osnovu [inlmath]10[/inlmath] drugu jednačinu:
[dispmath]\log_{10}x^y=\log_{10}0,01[/dispmath]
[dispmath]y\cdot\log_{10}x=-2\quad\left(1\right)[/dispmath]
Iz prve jednačine imamo:
[dispmath]\log_{10}x=1-y[/dispmath]
pa zamenimo u [inlmath]\left(1\right)[/inlmath]:
[dispmath]y\cdot\left(1-y\right)=-2[/dispmath]
[dispmath]-y^2+y+2=0[/dispmath]
[dispmath]y_{1,2}=\frac{-1\pm\sqrt{1+8}}{-2}[/dispmath]
[dispmath]y_{1,2}=\frac{-1\pm 3}{-2}[/dispmath]
[dispmath]y_1=-1,\quad y_2=2[/dispmath]
[dispmath]y_1=-1\quad\Rightarrow\quad\log_{10}x=1-y=2\quad\Rightarrow\quad x=100[/dispmath]
[dispmath]y_1=2\quad\Rightarrow\quad\log_{10}x=1-y=-1\quad\Rightarrow\quad x=\frac{1}{10}[/dispmath]
Oba rešenja po [inlmath]x[/inlmath] ispunjavaju uslov za definisanost, [inlmath]x>0[/inlmath], pa će sistem imati dva rešenja:
[dispmath]\enclose{box}{\begin{array}{ll}
\left(x,y\right)=\left(100,-1\right) \\
\left(x,y\right)=\left(\frac{1}{10},2\right)
\end{array}}[/dispmath]
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