[tex] \it 2^{x+2} + 3\cdot2^{x-1} = \dfrac{11}{4} \Rightarrow 2^2\cdot2^x +3\cdot2^{-1}\cdot2^x = \dfrac{11}{4} \Rightarrow 2^x(4 + 3\cdot \dfrac{1}{2}) = \dfrac{11}{4}\\ \\ \\ \Rightarrow 2^x(4+\dfrac{3}{2})=\dfrac{11}{4} \Rightarrow 2^x\cdot\dfrac{11}{2} = \dfrac{11}{4}|_{\cdot\frac{2}{11}} \Rightarrow 2^x = \dfrac{1}{2} \Rightarrow 2^x=2^{-1} \Rightarrow x = -1 [/tex]
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[tex] \it log_2(x+3) +log_2x = 2 \Rightarrow log_2(x+3)\cdot x = 2 \Rightarrow (x+3)\cdot x = 2^2 \Rightarrow \\ \\ \\ \Rightarrow (x+3)\cdot x = 4 = 4\cdot1 \Rightarrow x = 1 [/tex]
Verificare:
[tex] \it x=1 \Rightarrow log_2(1+3) +log_2 1 = 2 \Rightarrow log_24+0=2 \Rightarrow 2=2 \ (Adev\breve{a}rat) [/tex]
Deci, x=1 este soluție a ecuației inițiale.