Ecuatii exponentiale

annonimus Dum Feb 26, 2012 6:22 pm

Salut. Am nevoie de ajutor la 3 ecuatii exponentiale:

1) 4^|x-3| + 4^|x+1| = 4^x
2) 3^x + 4^x = 5^x
3) (2/5)^x = -3x^2 + 2x - 1
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Multumesc anticipat!

**Admin** Dum Feb 26, 2012 7:19 pm

annonimus a scris:2) 3^x + 4^x = 5^x

$\begin{array}{l} {3^x} + {4^x} = {5^x} \Leftrightarrow {\left( {\frac{3}{5}} \right)^x} + {\left( {\frac{4}{5}} \right)^x} = 1\\ {\text{Cum }}f\left( x \right) = {a^x}{\text{ functie descrescatoare (cazul nostru), bijectiva (nu voi demonstra acest lucru)}}\\ \Rightarrow {\text{ecuatia are o singura solutie :}}\\ {\text{Aceasta solutie se determina prin incercari :}}\\ {\text{luam }}x = 0 \Rightarrow {\left( {\frac{3}{5}} \right)^0} + {\left( {\frac{4}{5}} \right)^0} = 1 \Leftrightarrow 1 + 1 = 1\left( {fals} \right)\\ {\text{luam }}x = 2 \Rightarrow {\left( {\frac{3}{5}} \right)^2} + {\left( {\frac{4}{5}} \right)^2} = 1 \Leftrightarrow \frac{9}{{25}} + \frac{{16}}{{25}} = \frac{{25}}{{25}} = 1\left( {ok!!!} \right) \end{array}$

**Admin** Dum Feb 26, 2012 7:29 pm

annonimus a scris:3) (2/5)^x = -3x^2 + 2x - 1

$\begin{array}{l} {\left( {\frac{2}{5}} \right)^x} = - 3{x^2} + 2x - 1\\ f\left( x \right) = {\left( {\frac{2}{5}} \right)^x} \to {\text{Functie descrescatoare, pozitiva }}\left( {{\text{peste axa Ox}}} \right){\text{.}}\\ {\rm{g}}\left( x \right) = - 3{x^2} + 2x - 1 \to {\text{Functie descrescatoare, are varful (se calculeaza) sub axa Ox}}{\rm{.}}\\ f\left( x \right) = g\left( x \right) \to {\text{nu }}\exists {\text{ solutii}}{\rm{.}} \end{array}$

**Admin** Dum Feb 26, 2012 11:10 pm

annonimus a scris:1) 4^|x-3| + 4^|x+1| = 4^x

Soluția completa arata în felul următor :
$\begin{array}{l} {4^{\left| {x - 3} \right|}} + {4^{\left| {x + 1} \right|}} = {4^x}\\ \left| {x - 3} \right| = \left\{ \begin{array}{l} x - 3,\,\,x - 3 \ge 0\\ - \left( {x - 3} \right),\,\,x - 3 < 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x - 3,\,\,x \ge 3\\ - \left( {x - 3} \right),\,\,x < 3 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x - 3,\,\,x \in \left[ {3, + \infty } \right)\\ - \left( {x - 3} \right),\,\,\,\,x \in \left( { - \infty ,3} \right) \end{array} \right.\\ \left| {x + 1} \right| = \left\{ \begin{array}{l} x + 1,\,\,x + 1 \ge 0\\ - \left( {x + 1} \right),\,\,x + 1 < 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x + 1,\,\,x \ge - 1\\ - \left( {x + 1} \right),\,\,x < - 1 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x - 3,\,\,x \in \left[ { - 1, + \infty } \right)\\ - \left( {x - 3} \right),\,\,\,\,x \in \left( { - \infty , - 1} \right) \end{array} \right. \end{array}$

**Admin** Dum Feb 26, 2012 11:13 pm

Continuare problema 1 :
Ecuatii exponentiale 1-e1330269464456

**Admin** Dum Feb 26, 2012 11:19 pm

Continuare problema 1 :

Ecuatii exponentiale 2

**Admin** Dum Feb 26, 2012 11:21 pm

Dupa cum vedeți, nu avem soluții la problema 1...am scindat problema în mai multe parții din cauze tehnice sper ca ați înțeles metoda de rezolvare .

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