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andrei_93
Pe: 2010-08-31T16:59:52+03:00 In: In:

sisteme logaritmi

Am un exercitiu la care cred ca trebuie sa fac un artificiu de calcul….dar nu imi dau seama .
Mi se cere sa se rezolve sistemul.
log_{2}(x+1)+log_{3}(y+1)=2

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3 raspunsuri

  1. Sigma2
    maestru (V)

    Tu vorbesti de sistem, dar postezi o singura ecuatie.
    Aceasta duce la o nederminare (poti avea mai multe solutii). Ex
    x=0 y=8
    x=1 y=2
    x=3 y=0

  2. andrei_93

    Scuzele mele. Eram ff obosit . Sistemul arata asa:

    \left\{\begin {matrix} log_{2}(x+1)+log_{3}(y+1)=2\\ (x+1)(y+1)=6 \end{matrix}\right.

  4. sindex

        \[\begin{array}{l}<br/> 1.C.E.<E>:x</E>,y > - 1 \\ <br/> 2.\left\{ \begin{array}{l}<br/> {\log _2}(x + 1) + {\log _3}(y + 1) = 2 \\ <br/> (x + 1)(y + 1) = 6 \\ <br/> \end{array} \right. \\ <br/> 3.{\rm{Logaritmezi in baza 2 a II - a ecuatie:}} \\ <br/> (x + 1)(y + 1) = 6|{\log _2} \Rightarrow {\log _2}(x + 1) + {\log _2}(y + 1) = {\log _2}6{\rm{ (aplici proprietatea lg a + lg b = lg ab) [1]}} \\ <br/> 4.{\log _3}(y + 1) = \frac{{{{\log }_2}(y + 1)}}{{{{\log }_2}3}}{\rm{ (formula ptr schimbarea de baza)}} \\ <br/> \left. \begin{array}{l}<br/> 5.{\rm{Rescrii prima ecuatie tinand cont de punctul 4: }}{\log _2}(x + 1) + \frac{{{{\log }_2}(y + 1)}}{{{{\log }_2}3}} = 2 \\ <br/> {\rm{Din [1] }} \Rightarrow {\log _2}(y + 1) = {\log _2}6 - {\log _2}(x + 1) \\ <br/> \end{array} \right| \Leftrightarrow {\log _2}(x + 1) + \frac{{{{\log }_2}6 - {{\log }_2}(x + 1)}}{{{{\log }_2}3}} = 2 \\ <br/> 7.{\log _2}(x + 1) + \frac{{{{\log }_2}6 - {{\log }_2}(x + 1)}}{{{{\log }_2}3}} = 2 \Leftrightarrow {\log _2}3 \cdot {\log _2}(x + 1) + {\log _2}6 - {\log _2}(x + 1) = 2{\log _2}3 \\ <br/> 8.{\log _2}(x + 1) = t \Rightarrow t \cdot {\log _2}3 + {\log _2}6 - t = 2{\log _2}3 \\ <br/> t({\log _2}3 - 1) + {\log _2}6 = 2{\log _2}3 \Rightarrow t = \frac{{2{{\log }_2}3 - {{\log }_2}6}}{{{{\log }_2}3 - 1}} = \underbrace {\frac{{2{{\log }_2}3 - {{\log }_2}2 - {{\log }_2}3}}{{{{\log }_2}3 - 1}}}_{{\rm{(tii cont ca lo}}{{\rm{g}}_2}6 = {{\log }_2}2 + {{\log }_2}3)} = \frac{{{{\log }_2}3 - 1}}{{{{\log }_2}3 - 1}} = 1 \\ <br/> \left. \begin{array}{l}<br/> t = 1 \\ <br/> t = {\log _2}(x + 1) \\ <br/> \end{array} \right| \Rightarrow x + 1 = 2 \Rightarrow x = 1 \\ <br/> {\rm{9}}{\rm{.Intrii in oricare ecuatie a sistemului: spre exemplu}}{\rm{, prima}} \\ <br/> \left. \begin{array}{l}<br/> {\log _2}(x + 1) + {\log _3}(y + 1) = 2 \\ <br/> x = 1 \\ <br/> \end{array} \right| \Rightarrow 1 + {\log _3}(y + 1) = 2 \Rightarrow y = 2 \\ <br/> {\rm{10}}{\rm{.Verifici in a doua:}} \\ <br/> \left. \begin{array}{l}<br/> {\rm{x = 1}} \\ <br/> {\rm{y = 2}} \\ <br/> \end{array} \right| \Rightarrow 2 \cdot 3 = 6(A){\rm{ deci S = \{ (1}}{\rm{,2)\} }} \\ <br/> \end{array}\]

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