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Care este soluția sistemului: [tex] \left \{ {{ 2^{x} 5^{y}=250 } \atop { 2^{y} 5^{x}=40 }} \right. [/tex]

Răspuns :

AcelOm
[tex]2^x\cdot5^y\cdot2^y\cdot5^x=250\cdot40[/tex]
[tex]10^x\cdot10^y=10000[/tex]
[tex]10^{x+y}=10000[/tex]
[tex]x+y=4[/tex]

[tex]2^x\mid250[/tex]
[tex]2\mid250,~dar~2^2\nmid250[/tex]
[tex]\Rightarrow2^x=2\Rightarrow x=1[/tex]
[tex]\Rightarrow y=3[/tex]
[tex]2^1\cdot5^3=2\cdot125=250[/tex]
[tex]2^3\cdot5^1=8\cdot5=40[/tex]

[tex]\boxed{x=1,~y=3}[/tex]
Rayzen
[tex]\left\{ \begin{array}{c} 2^x\cdot 5^y = 250 \\ 2^y\cdot 5^x = 40 \end{array} \right | \\ \\ \boxed{1} \quad 2^x = \dfrac{250}{5^y} \Rightarrow x = log_\big2 \dfrac{250}{5^y} \Rightarrow x = log_\big2250 - log_\big2 5^y \Rightarrow \\ \Rightarrow x = log_\big2250 - ylog_\big2 5 \\ \\ \boxed{2} \quad 5^x = \dfrac{40}{2^y} \Rightarrow x = log_\big5 \dfrac{40}{2^y} \Rightarrow x = log_\big5 40 - log_\big5 2^y \Rightarrow \\ \Rightarrow x = log_\big5 40 - ylog_\big5 2 [/tex]


[tex]\\ $Egalam: \\ \\ $ log_\big2250 - ylog_\big2 5=log_\big5 40 - ylog_\big5 2 \Rightarrow \\ \\ \Rightarrow ylog_\big5 2 - ylog_\big25 = log_\big5 40 - log_\big2250 \Rightarrow \\ \\ \Rightarrow y \cdot(log_\big5 2 - log_\big 2 5) = log_\big 5 40 - log_\big 2 250 \Rightarrow \\ \\ \Rightarrow y = \dfrac{ log_\big 5 40 - log_\big 2 250}{log_\big5 2 - log_\big 2 5} \Rightarrow y = \dfrac{log_\big5 2^3\cdot 5- log_\big 2 5^3\cdot 2}{log_\big5 2 - log_\big 2 5} \Rightarrow \\ \\[/tex]


[tex] \Rightarrow y = \dfrac{log_\big5 2^3+ log_\big 55- (log_\big 2 5^3+log_\big 2 2)}{log_\big5 2 - log_\big 2 5} \Rightarrow \\ \\ \Rightarrow y =\dfrac{3log_\big5 2+ log_\big 55- 3log_\big 2 5-log_\big 2 2}{log_\big5 2 - log_\big 2 5} \Rightarrow \\ \\ \Rightarrow y =\dfrac{3\cdot(log_\big5 2 - log_\big 2 5) + log_\big 5 5 - log_\big2 2}{log_\big5 2 - log_\big 2 5} \Rightarrow [/tex]


[tex] \\ \\ \Rightarrow y=\dfrac{3\cdot(log_\big5 2 - log_\big 2 5) + 1 - 1}{log_\big5 2 - log_\big 2 5} \Rightarrow y=\dfrac{3\cdot(log_\big5 2 - log_\big 2 5)}{log_\big5 2 - log_\big 2 5} \Rightarrow \boxed{y=3} \\ \\ y=3 \Rightarrow 2^x\cdot 5^3 = 250 \Rightarrow 2^x\cdot 5^3 = 125\cdot 2 \Rightarrow 2^x \cdot 5^3 = 5^3\cdot 2 \Rightarrow \\ \\ \Rightarrow2^x= \dfrac{5^3\cdot 2}{5^3} \Rightarrow 2^x = 2 \Rightarrow \boxed{x = 1}[/tex]

[tex]\Rightarrow \boxed{(x,y) = (1,3)}[/tex]