Rezolvati ecuatia
\( \log_2(x+2)\cdot\log_x(3^x+7)=x^3 \)
Ecuatie logaritmica 2
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Marius Mainea
- Gauss
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andy crisan
- Pitagora
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Daca \( x\in (0,1) \), atunci \( \log_{2}(x+2)>0 \) si \( \log_{x}(3^x+7)<0 \) si \( x^3>0 \), deci nu avem solutii in \( (0,1) \).
Daca \( x\in(1,2) \), atunci \( 3^x+7>4^x \) (se arata imediat). Cum \( x>1 \) rezulta \( \log_{x}(3^x+7)>\log_{x}4^x=2x\log_{x}2 \) (1)
Cum \( x\in(1,2)\Rightarrow x^2(x-1)<4 \Leftrightarrow x^2(x-1)(x+1)<4(x+1)\Leftrightarrow x^4<(x+2)^2 \) dar \( x^4>x^{x^2}\Rightarrow x^{x^2}<(x+2)^2\Leftrightarrow \log_{2}x^{x^2}<\log_{2}(x+2)^2\Leftrightarrow \frac{x^{2}\log_{2}x}{2}<\log_{2}(x+2) \)(2).
Inmultind relatiile (1) si (2) obtinem ca \( \log_{2}(x+2)\log_{x}(3^x+7)>x^3 \).
Analog pt \( x>2 \).
Rezulta ca \( x=2 \) e unica solutie.
Daca \( x\in(1,2) \), atunci \( 3^x+7>4^x \) (se arata imediat). Cum \( x>1 \) rezulta \( \log_{x}(3^x+7)>\log_{x}4^x=2x\log_{x}2 \) (1)
Cum \( x\in(1,2)\Rightarrow x^2(x-1)<4 \Leftrightarrow x^2(x-1)(x+1)<4(x+1)\Leftrightarrow x^4<(x+2)^2 \) dar \( x^4>x^{x^2}\Rightarrow x^{x^2}<(x+2)^2\Leftrightarrow \log_{2}x^{x^2}<\log_{2}(x+2)^2\Leftrightarrow \frac{x^{2}\log_{2}x}{2}<\log_{2}(x+2) \)(2).
Inmultind relatiile (1) si (2) obtinem ca \( \log_{2}(x+2)\log_{x}(3^x+7)>x^3 \).
Analog pt \( x>2 \).
Rezulta ca \( x=2 \) e unica solutie.