Question

Heeei,as avea nevoie de ajutor la acest exercitiu.
${4}^{x} \times {9}^{ \frac{1}{x} } + {4}^{ \frac{1}{x} } \times {9}^{x} = 210$

Answer 1

$\displaystyle Evident~nu~putem~avea~x=0. \\ \\Daca~x<0~atunci~4^x \cdot 9^{\frac{1}{x}}+4^{\frac{1}{x}} \cdot 9^x<1+1=2. \\ \\ Deci~x>0. \\ \\ Scriindu-l~pe~9~drept~4^{\log_4 9}~vom~avea: \\ \\ 4^x \cdot 9^{ \frac{1}{x}}=4^x \cdot 4^{ \frac{\log_49}{x}=4^{x+ \frac{\log_49}{x}}}; \\ \\ 4^\frac{1}{x} \cdot 9^x=4^{\frac{1}{x}} \cdot 4^{(\log_49)x}=4^{(\log_49)x+ \frac{1}{x}}.$

$\displaystyle Functiile~de~forma~ax+ \frac{b}{x}~sunt~(strict)~convexe~pe~(0, \infty). \\ \\ (asta~pentru~ca~ax~si~\frac{b}{x}~sunt~(strict)~convexe~pe~(0, \infty).) \\ \\ O~teorema~ne~spune~ca~daca~f~si~g~sunt~convexe,~iar~f~este~(strict) \\ \\ crescatoare,~atunci~f \circ g~este~(strict)~convexa. \\ \\ Cum~4^x~este~strict~crescatoare,~iar~functiile~x+ \frac{\log_49}{x}~si~(\log_49)x+ \frac{1}{x}$

$\displaystyle sunt~convexe,~rezulta~ca~4^{x+ \frac{\log_49}{x}} ~si~4^{(\log_49)x+ \frac{1}{x}}~sunt~strict~convexe \\ \\ pe ~(0, \infty). \\ \\ Deci~4^x \cdot 9^{\frac{1}{x}]+4^{\frac{1}{x}}} \cdot 9^x~este~strict~convexa~pe~(0, \infty),~deci~ecuatia~data \\ \\ are~cel~mult~doua~solutii,~si~cum~2~si~ \frac{1}{2}~sunt~solutii,~rezulta~ca \\ \\ acestea~sunt~singurele~solutii~ale~ecuatiei.$