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Fie \( a\in(0,1)\cup(1,\infty) \). Rezolvati ecuatia
\( \log_x (x+1)=\log_a (a+1) \).

1) Daca a>1, atunci \( f : (1,+\infty) \rightarrow\mathbb{R} \) \( f(x)=\frac{\ln(x+1)}{\ln(x)} \) este stict descrescatoare, deci x=a solutie unica.

2) Daca a<1, notam \( b=\frac{1}{a} \), \( y=\frac{1}{x} \) si reducem la cazul precedent.

Pentru \( x\in\(0,1\) \):

Functia \( \log_x(x+1) \) este strict descrescatoare pe ambele intervale \( \(0,1\) \) si \( \(1,\infty) \), iar \( \log_x(x+1)<0 \) daca \( x<1 \) si \( \log_x(x+1)>0 \) daca \( x>1 \), deci ecuatia are solutie unica \( a \).

\( \log_x (x+1)=\log_a (a+1)\Rightarrow\frac{\log_a (x+1)}{\log_a x}=\log_a (a+1). \) Notam \( \log_a x=y\Rightarrow\frac{\log_a (a^y+1)}{y}=\log_a (a+1)\Rightarrow\frac{\log_a (a^y+1)}{\log_a (a+1)}=y\Rightarrow \)
\( \Rightarrow\log_{(a+1)} (a^y+1)=y\Rightarrow(a+1)^y=a^y+1\Rightarrow(\frac{a}{a+1})^y+(\frac{1}{a+1})^y=1\Rightarrow \) cu solutia unica \( y=1\Rightarrow x=a \) este solutie unica a ecuatiei.