Question

Determinati in fiecare dintre urmatoarele cazuri nr real x pentru care egalitatea este adevarata

Answer 1

   
$\displaystyle\\ \text{Folosim formula:}\\\\\log_{a^p}(x)=\frac{1}{p}\cdot\log_a(x)\\\\ \text{Rezolvare:}\\\\ \text{Prima ecuatie:}\\\\ \log_2(x)+\log_4(x)=6\\\\ \log_2(x)+\log_{2^2}(x)=6\\\\ \log_2(x)+\frac{1}{2}\log_2(x)=6~~ \text{Dam factor comun pe }~\log_2(x)\\\\ \log_2(x)\Big(1+\frac{1}{2}\Big)=6\\\\ \frac{3}{2}\cdot\log_2(x)=6\\\\ \log_2(x)=6\times\frac{2}{3}\\\\ \log_2(x)=4\\\\ x=2^4=\boxed{\bf 16}$


[tex]\displaystyle\\ \text{A doua ecuatie:}\\\\ \log_3(x)+\log_9(x)=3\\\\ \log_3(x)+\log_{3^2}(x)=3\\\\ \log_3(x)+ \frac{1}{2} \cdot\log_3(x)=3~~~\text{Dam factor comun pe }\log_3(x)\\\\ \log_3(x)\cdot \Big(1+\frac{1}{2} \Big)=3\\\\ \frac{3}{2} \cdot \log_3(x) = 3\\\\ \log_3(x)= 3\cdot \frac{2}{3} \\\\ \log_3(x)= 2\\\\ x = 3^2 = \boxed{\bf 9} [/tex]


[tex]\displaystyle\\ \text{A treia ecuatie:}\\\\ \log_4(x)-\log_2(x)=0\\\\ \log_{2^2}(x)-\log_2(x)=0\\\\ \frac{1}{2} \cdot \log_2(x)-\log_2(x)=0~~~\text{Dam factor pe } \log_2(x)\\\\ \log_2(x)\cdot\Big( \frac{1}{2}-1\Big)=0\\\\ -\frac{1}{2}\cdot\log_2(x)=0\\\\ \log_2(x) = \frac{0}{-\dfrac{1}{2}} \\\\ \log_2(x)=0 x = 2^0 = \boxed{\bf 1}[/tex]