Question

log_{9} (4^{x}+1)=log_{4}(9^{x}-1)
x=?

Multumesc !

Answer 1

$\displaystyle Problemele~de~tipul~acesta~pot~fi~rezolvate~prin~doua~metode. \\ \\ Metoda~1:~Observam~ca~expresiile~\log_9(4^x+1)~si~\log_4(9^x-1) \\ \\ au~sens~pentru~orice~x \in \mathbb{R}~deci~putem~considera~functiile \\ \\ f: \mathbb{R} \to \mathbb{R},~f(x)= \log_9(4^x+1)~si~g: \mathbb{R} \to \mathbb{R},~g(x)= \log_4(9^x-1). \\ \\ Ambele~functii~sunt~strict~crescatoare,~deci~injective. \\ \\ Este~usor~de~demonstrat~ca~f~si~g~sunt~surjective.$

$\displaystyle Scurta~demonstratie:~Pentru~orice~t \in \mathbb{R},~ecuatia~f(x)=t \\ \\ admite~solutia~\log_4(9^t-1).~De~asemenea~pentru~orice~t \in \mathbb{R}, \\ \\ ecuatia~g(x)=t~admite~solutia~\log_9(4^t+1). \\ \\ Asadar~f~si~g~sunt~bijective. \\ \\ Abia~acum~incepe~rezolvarea~efectiva~a~ecuatiei. \\ \\ Avem~f(g(x))=x~\forall ~x \in \mathbb{R},~deci~functiile~f~si~g~sunt~inverse. \\ \\ (Precizez~ca~acest~lucru~se~putea~verifica~eventual~calculand \\ \\ inversa~lui~f.)$

$\displaystyle Deci~ecuatia~noastra~este~f(x)=f^{-1}(x).~Se~stie~insa~ca \\ \\ graficul~undei~functii~si~graficul~functiei~sale~inverse~sunt \\ \\ simetrice~fata~de~prima~bisectoare. \\ \\ Prin~urmare~egalitatea~f(x)=f^{-1}(x)~implica~f(x)=x.$

$\displaystyle Deci~\log_9(4^x+1)=x \Leftrightarrow 4^x+1=9^x \Leftrightarrow \left( \frac{4}{9} \right)^x+ \left( \frac{1}{9} \right)^x=1,~cu~ \\ \\ solutia~unica~ \frac{1}{2}~(pentru~ca~ membrul~stang~este~strict~descrescator). \\ \\ Metoda~2:~Se~noteaza~ \log_9(4^x+1)= \log_4(9^x-1)=a. \\ \\ Se~obtine~4^x+1=9^a~si~9^x-1=4^a.$

$\displaystyle Se~aduna~aceste~relatii:~4^x+9^x=4^a+9^a. \\ \\ Se~considera~functia~f: \mathbb{R} \to \mathbb{R},~f(x)=4^x+9^x. \\ \\ Functia~f~este~strict~crescatoare~(suma~de~doua~functii~strict \\ \\ crescatoare). \\ \\ Deci~ecuatia~f(x)=f(a)~admite~solutie~unica~x=a. \\ \\ Rezulta~\log_9(4^x+1)=x,~iar~finalizarea~se~face~ca~la~prima \\ \\ metoda.$