Question

Cum se rezolvă exercițiul: 4 la x-7 ori 10 la x + 10 ori 25 la x=0
${4}^{x} - 7 \times {10}^{x} + 10 \times {25}^{x} = 0$
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Answer 1

Răspuns:

$4^{x}-7*10^{x}+10*25^{x}=0,~ecuatie~omogena\\(2^{x})^{2}-7*2^{x}*5^{x}+10*(5^{x})^{2}=0~|:(5^{x})^{2},~~\dfrac{(2^{x})^{2}}{(5^{x})^{2}}-7*\dfrac{2^{x}*5^{x}}{(5^{x})^{2}}+10*\dfrac{(5^{x})^{2}}{(5^{x})^{2}}=0,~\\ ((\dfrac{2}{5})^{x})^{2}-7*(\dfrac{2}{5})^{x}+10=0.~~Fie~(\dfrac{2}{5})^{x}=t,~obtinem\\t^{2}-7t+10=0,~delta=49-40=9>0,~deci~t_{1}=\dfrac{7-3}{2} =2,~~t_{2}=\dfrac{7+3}{2} =5.~~Revenim~la~substitutie:(\dfrac{2}{5})^{x}=2,~deci~x=log_{\frac{2}{5} }2,~si~$

$(\dfrac{2}{5})^{x}=5,~deci~x=log_{\frac{2}{5} }5$

Explicație pas cu pas:

Answer 2

$4^x-7\cdot 10^x+10\cdot 25^x=0\\\\\Rightarrow 2^{2x}-7\cdot (2\cdot 5)^x+10\cdot 5^{2x} = 0\Big|:(5^{2x}\neq 0)\\ \\ \Rightarrow \left(\dfrac{2}{5}\right)^{2x}-7\cdot \left(\dfrac{2}{5}\right)^x+10 = 0\\ \\\text{Notez: } \left(\dfrac{2}{5}\right)^{x} = t\\ \\ \Rightarrow t^2-7t+10 = 0\\ \Rightarrow t^2-5t-2t+10 = 0\\\Rightarrow t(t-5)-2(t-5) = 0\\ \Rightarrow (t-5)(t-2) = 0$

$\bullet \,\,\,\,t = 5 \Rightarrow \left(\dfrac{2}{5}\right)^x = 5 \Rightarrow x = \log_{\frac{2}{5}}5\\ \\ \bullet \,\,\,\,t = 2 \Rightarrow \left(\dfrac{2}{5}\right)^x = 2 \Rightarrow x = \log_{\frac{2}{5}}2\\\\\Rightarrow \boxed{S =\left\{\log_{\frac{2}{5}}5;\,\log_{\frac{2}{5}}2\right\}}$