Ako je [inlmath]4^x+4^{-x}=38[/inlmath], onda je [inlmath]2^x+2^{-x}=[/inlmath]
Probala sam sa smenom [inlmath]t=4^x[/inlmath] i ne ide.

Probaj sa smenom [inlmath]2^x=t[/inlmath]

Može i [inlmath]4^x=t[/inlmath]...[dispmath]4^x+4^{-x}=38\quad /\cdot 4^x[/dispmath][dispmath]\left(4^x\right)^2+1-38\cdot 4^x=0[/dispmath][dispmath]t^2-38t+1=0[/dispmath][dispmath]t_{1,2}=\frac{38\pm\sqrt{38^2-4}}{2}=19\pm\sqrt{19^2-1}=19\pm\sqrt{360}=19\pm 6\sqrt{10}[/dispmath][dispmath]x=\log_4 t\quad\Rightarrow\quad x=\log_4\left(19\pm 6\sqrt{10}\right)[/dispmath][dispmath]2^x+2^{-x}=2^{\log_4\left(19\pm 6\sqrt{10}\right)}+2^{-\log_4\left(19\pm 6\sqrt{10}\right)}[/dispmath][dispmath]2^x+2^{-x}=\left(4^\frac{1}{2}\right)^{\log_4\left(19\pm 6\sqrt{10}\right)}+\left(4^\frac{1}{2}\right)^{-\log_4\left(19\pm 6\sqrt{10}\right)}[/dispmath][dispmath]2^x+2^{-x}=4^{\frac{1}{2}\log_4\left(19\pm 6\sqrt{10}\right)}+4^{-\frac{1}{2}\log_4\left(19\pm 6\sqrt{10}\right)}[/dispmath][dispmath]2^x+2^{-x}=4^{\log_4\sqrt{19\pm 6\sqrt{10}}}+4^{\log_4\frac{1}{\sqrt{19\pm 6\sqrt{10}}}}[/dispmath][dispmath]2^x+2^{-x}=\sqrt{19\pm 6\sqrt{10}}+\frac{1}{\sqrt{19\pm 6\sqrt{10}}}[/dispmath][dispmath]2^x+2^{-x}=\frac{19\pm 6\sqrt{10}+1}{\sqrt{19\pm 6\sqrt{10}}}[/dispmath][dispmath]2^x+2^{-x}=\frac{20\pm 6\sqrt{10}}{\sqrt{19\pm 6\sqrt{10}}}[/dispmath]Radi oslobađanja od korena u imeniocu, prikažimo potkorenu veličinu kao kvadrat binoma:
[inlmath]19\pm 6\sqrt{10}=\left(a\pm b\right)^2=a^2+b^2\pm 2ab[/inlmath]
[inlmath]a^2+b^2=19[/inlmath]
[inlmath]2ab=6\sqrt{10}\quad /:2[/inlmath]
[inlmath]ab=3\sqrt{10}\quad\Rightarrow\quad b=\frac{3\sqrt{10}}{a}\quad\Rightarrow\quad b^2=\frac{90}{a^2}[/inlmath]
[inlmath]a^2+\frac{90}{a^2}=19\quad /\cdot a^2[/inlmath]
[inlmath]\left(a^2\right)^2-19a^2+90=0[/inlmath]
[inlmath]\left(a^2\right)_{1,2}=\frac{19\pm\cancelto{1}{\sqrt{19^2-4\cdot 90}}}{2}[/inlmath]
[inlmath]a=\pm 3\quad\lor\quad a=\pm\sqrt{10}[/inlmath]
[inlmath]a=3\quad\Rightarrow\quad b=\frac{\cancel 3\sqrt{10}}{\cancel 3}=\sqrt{10}[/inlmath]
[inlmath]19\pm 6\sqrt{10}=\left(3\pm\sqrt{10}\right)^2[/inlmath][dispmath]2^x+2^{-x}=\frac{20\pm 6\sqrt{10}}{\sqrt{\left(3\pm\sqrt{10}\right)^2}}[/dispmath][dispmath]2^x+2^{-x}=\frac{20\pm 6\sqrt{10}}{\left|3\pm\sqrt{10}\right|}[/dispmath][inlmath]3-\sqrt{10}<0,\quad 3+\sqrt{10}>0\quad\Rightarrow\quad\left|3\pm\sqrt{10}\right|=\sqrt{10}\pm 3[/inlmath][dispmath]2^x+2^{-x}=\frac{20\pm 6\sqrt{10}}{\sqrt{10}\pm 3}\cdot\frac{\sqrt{10}\mp 3}{\sqrt{10}\mp 3}[/dispmath][dispmath]2^x+2^{-x}=\frac{\left(20\pm 6\sqrt{10}\right)\left(\sqrt{10}\mp 3\right)}{10-9}[/dispmath][dispmath]2^x+2^{-x}=\left(20\pm 6\sqrt{10}\right)\left(\sqrt{10}\mp 3\right)[/dispmath][dispmath]2^x+2^{-x}=20\sqrt{10}\pm\cancel{60}\mp\cancel{60}-18\sqrt{10}[/dispmath][dispmath]\enclose{box}{2^x+2^{-x}=2\sqrt{10}}[/dispmath]