Sa se rezolve ecuatia:
\( 2^{x-1}+2^{\frac{1}{\sqrt{x}}}=3. \)
Marius Cavachi, G.M. 1988
Ecuatie exponentiala simpla
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Cezar Lupu
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Ecuatie exponentiala simpla
An infinite number of mathematicians walk into a bar. The first one orders a beer. The second orders half a beer. The third, a quarter of a beer. The bartender says “You’re all idiots”, and pours two beers.
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Mateescu Constantin
- Newton
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Conditia de existenta impune \( x>0 \).
Inmultind cu \( 2 \) egalitatea obtinem \( 2^x+2\cdot 2^{\frac{1}{\sqrt x}}=6. \)
Dar din \( AM-GM \) avem \( 2^x+2\cdot 2^{\frac{1}{sqrt x}}\ \ge\ 3\sqrt[3]{2^x\cdot 2^{\frac{1}{\sqrt x}}\cdot 2^{\frac{1}{\sqrt x}}}=3\sqrt[3]{2^{x+\frac{2}{\sqrt x}}}\ \ge\ 3\sqrt[3]{2^3}=6 \), cu egalitate cand \( x=\frac{1}{\sqrt x} \), adica pentru \( x=1 \).
Fiind chiar in cazul de egalitate rezulta ca \( x=1 \) este solutie unica.
Inmultind cu \( 2 \) egalitatea obtinem \( 2^x+2\cdot 2^{\frac{1}{\sqrt x}}=6. \)
Dar din \( AM-GM \) avem \( 2^x+2\cdot 2^{\frac{1}{sqrt x}}\ \ge\ 3\sqrt[3]{2^x\cdot 2^{\frac{1}{\sqrt x}}\cdot 2^{\frac{1}{\sqrt x}}}=3\sqrt[3]{2^{x+\frac{2}{\sqrt x}}}\ \ge\ 3\sqrt[3]{2^3}=6 \), cu egalitate cand \( x=\frac{1}{\sqrt x} \), adica pentru \( x=1 \).
Fiind chiar in cazul de egalitate rezulta ca \( x=1 \) este solutie unica.
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DrAGos Calinescu
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