Question

1. sa se calculeze combinari de 3 luate cate 2+3!
2. sa se determine solutiile reale ale ecuatiei log5(3x+4)=2

Answer 1

$a) C_{n}^k= \frac{n!}{k!(n-k)!},deci: C_{3}^2+3!= \frac{3!}{2!1!}+1*2*3= \frac{1*2*3}{1*2*1}+6=3+6=9$ b) log_{5}(3x+4)=2, Conditia: 3x+4>0, adica x>-4/3. Ecuntia devine 5^2=3x+4, deci 25=3x+4. 3x=21, sau x=7 solutie corespunzatoare conditiei. [/tex]

Answer 2

$\displaystyle 1).C_3^2+3!= \frac{3!}{(3-2)! \cdot 2!} +1 \cdot 2 \cdot 3= \frac{3!}{1! \cdot 2!} +6= \frac{\not2! \cdot 3}{1! \cdot \not2!} +6= \\ \\ = \frac{3}{1} +6=3+6=\boxed{9} \\ \\ 2).log_5(3x+4)=2 \\ \\ 3x+4\ \textgreater \ 0 \Rightarrow x\ \textgreater \ - \frac{4}{3} \Rightarrow x \in \left(- \frac{4}{3} , \infty \right) \\ \\ log_5(3x+4)=log_525 \\ \\ 3x+4=25 \Rightarrow 3x=25-4\Rightarrow 3x=21 \Rightarrow x= \frac{21}{3} \Rightarrow \boxed{x=7}$