منتدى الرياضيات رمز

بواسطة **smartboy** » الاثنين مارس 19, 2012 1:40 am

للتذكير فقط
hello:

بواسطة **caesar** » الاثنين مارس 19, 2012 7:13 am

السلام عليكم

$\begin{array}{l} \log _8^x + \;\log _4^{y^2 } \; = \;5\;\quad \Rightarrow \\ \\ \log _8^x + \;2\log _4^y \; = \;5\;\quad \Rightarrow \\ \\ \frac{{\log _{}^x }}{{\log _{}^8 }} + \;2\frac{{\log _{}^y }}{{\log _{}^4 }}\; = \;5\; \Rightarrow \\ \\ \frac{{\log _{}^x }}{{3\log _{}^2 }} + \;2\frac{{\log _{}^y }}{{2\log _{}^2 }}\; = \;5\; \Rightarrow \\ \\ \frac{{\log _{}^x + 3\log _{}^y }}{{3\log _{}^2 }}\;\; = \;5\; \Rightarrow \\ \\ \log _{}^{xy^3 } \;\; = \;5 \times 3\log _{}^2 \; \Rightarrow \\ \\ xy^3 \;\; = \;2^{15} \; \Rightarrow \\ \\ \log _8^y + \;\log _4^{x^2 } \; = \;7\;\quad \Rightarrow \\ \\ \frac{{\log _{}^y }}{{\log _{}^8 }} + \;2\frac{{\log _{}^x }}{{\log _{}^4 }}\; = \;7\;\quad \Rightarrow \\ \\ \frac{{\log _{}^y }}{{3\log _{}^2 }} + \;\frac{{\log _{}^x }}{{\log _{}^2 }}\; = \;7\;\quad \Rightarrow \\ \\ \frac{{\log _{}^y + 3\log _{}^x }}{{3\log _{}^2 }}\;\; = \;7\;\quad \Rightarrow \\ \\ \log _{}^y + 3\log _{}^x \;\; = \;21 \times \log _{}^2 \;\quad \Rightarrow \\ \\ x^3 y = 2^{21} \Rightarrow \\ \\ \frac{{x^3 y}}{{xy^3 \;\;\;}} = \frac{{2^{21} }}{{2^{15} }} \Rightarrow \\ \\ \left( {\frac{x}{{y\;\;\;}}} \right)^2 = 2^6 \Rightarrow x = 8y \Rightarrow \\ \\ x^3 y = 2^{21} \Rightarrow 2^9 y^4 = 2^{21} \Rightarrow \\ \\ y^4 = 2^{12} \Rightarrow y = \ 8 \Rightarrow x = \ 64 \\ \end{array}$

بواسطة **smartboy** » الاثنين مارس 19, 2012 4:18 pm

.......حقيقة حل رائع ومميز ....من رياضياتى مبدع ومميز.....وهذه محاولتى:

$\begin{array}{l} {\log _8}(x) + {\log _4}({y^2}) = 5...........(1)\\ {\log _8}(y) + {\log _4}({x^2}) = 7...........(2)\\ (1) + (2):\\ {\log _8}(xy) + {\log _{{2^2}}}({x^2}{y^2}) = 12\\ we.know.that:{\log _{{2^2}}}({x^2}{y^2}) = {\log _2}\left( {xy} \right)\\ \Rightarrow {\log _8}(xy) + {\log _2}(xy) = 12\\ let:{\log _8}(xy) = m\\ xy = {2^{3m}} \Rightarrow {\log _2}\left( {xy} \right) = 3m\\ m + 3m = 12 \Rightarrow m = 3 \Rightarrow \left[ {xy = 512} \right].............(3)\\ (2) - (1):\\ {\log _8}\left( {\frac{y}{x}} \right) + {\log _4}{\left( {\frac{x}{y}} \right)^2} = 2\\ {\log _8}\left( {\frac{y}{x}} \right) + {\log _2}\left( {\frac{x}{y}} \right) = 2\\ let:{\log _8}\left( {\frac{y}{x}} \right) = n \Rightarrow {2^{3n}} = \frac{y}{x} \Rightarrow {\log _2}\left( {\frac{x}{y}} \right) = - 3n\\ n - 3n = 2 \Rightarrow n = - 1\\ \Rightarrow \left[ {\frac{y}{x} = \frac{1}{8}} \right]............(4)\\ from:(3),(4):\\ x.\frac{x}{8} = 512 \Rightarrow x = \pm 64 \Rightarrow \left( {the.negative.sign.is.refused} \right)\\ \Rightarrow x = 64,y = 8 \end{array}$

بواسطة **caesar** » الاثنين مارس 19, 2012 5:16 pm

السلام عليكم
حلكم أدناه يدل على إنكم اسم على مسمى
very very smart sol. thanks alot
بارك الله فيكم

smartboy كتب:gg: .......حقيقة حل رائع ومميز ....من رياضياتى مبدع ومميز.....وهذه محاولتى:

$\begin{array}{l} {\log _8}(x) + {\log _4}({y^2}) = 5...........(1)\\ {\log _8}(y) + {\log _4}({x^2}) = 7...........(2)\\ (1) + (2):\\ {\log _8}(xy) + {\log _{{2^2}}}({x^2}{y^2}) = 12\\ we.know.that:{\log _{{2^2}}}({x^2}{y^2}) = {\log _2}\left( {xy} \right)\\ \Rightarrow {\log _8}(xy) + {\log _2}(xy) = 12\\ let:{\log _8}(xy) = m\\ xy = {2^{3m}} \Rightarrow {\log _2}\left( {xy} \right) = 3m\\ m + 3m = 12 \Rightarrow m = 3 \Rightarrow \left[ {xy = 512} \right].............(3)\\ (2) - (1):\\ {\log _8}\left( {\frac{y}{x}} \right) + {\log _4}{\left( {\frac{x}{y}} \right)^2} = 2\\ {\log _8}\left( {\frac{y}{x}} \right) + {\log _2}\left( {\frac{x}{y}} \right) = 2\\ let:{\log _8}\left( {\frac{y}{x}} \right) = n \Rightarrow {2^{3n}} = \frac{y}{x} \Rightarrow {\log _2}\left( {\frac{x}{y}} \right) = - 3n\\ n - 3n = 2 \Rightarrow n = - 1\\ \Rightarrow \left[ {\frac{y}{x} = \frac{1}{8}} \right]............(4)\\ from:(3),(4):\\ x.\frac{x}{8} = 512 \Rightarrow x = \pm 64 \Rightarrow \left( {the.negative.sign.is.refused} \right)\\ \Rightarrow x = 64,y = 8 \end{array}$

بواسطة **smartboy** » الثلاثاء مارس 20, 2012 3:53 pm

جزاك الله خيرا .......وهذه محاولة اخرى:

$\begin{array}{l} \Rightarrow we.know.that:{\log _{{2^2}}}({y^2}) = {\log _2}(y),{\log _{{2^2}}}({x^2}) = {\log _2}(x)\\ {\log _8}(x) + {\log _2}(y) = 5\\ {\log _8}(y) + {\log _2}(x) = 7\\ let:{\log _8}(x) = m \Rightarrow {2^{3m}} = x \Rightarrow {\log _2}(x) = 3m\\ ,let:{\log _8}(y) = n \Rightarrow {2^{3n}} = y \Rightarrow {\log _2}(y) = 3n\\ \Rightarrow m + 3n = 5\\ \Rightarrow n + 3m = 7\\ solving.the.two.equations:\\ m = 2,n = 1\\ \Rightarrow x = 64,y = 8 \end{array}$

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