Question

Rozwiąż równanie: [(81 ³√3)/√(27)]^6x = 3^x2 na zdjęciu jak to powinno wyglądać xD

Answer 1

[tex]\left(\dfrac{81\sqrt[3]{3}}{\sqrt{27}}\right)^{6x}=3^{x^2}\\\\\\\left(\dfrac{3^4\cdot3^{\frac{1}{3}}}{\sqrt{3^3}}\right)^{6x}=3^{x^2}\\\\\\\left(\dfrac{3^{4+\frac{1}{3}}}{3^{\frac{3}{2}}}\right)^{6x}=3^{x^2}\\\\\\\left(\dfrac{3^{4\frac{1}{3}}}{3^{\frac{3}{2}}}\right)^{6x}=3^{x^2}\\\\\\\left(\dfrac{3^{\frac{13}{3}}}{3^{\frac{3}{2}}}\right)^{6x}=3^{x^2}\\\\\\(3^{\frac{13}{3}-\frac{3}{2}})^{6x}=3^{x^2}\\\\(3^{\frac{26}{6}-\frac{9}{6}})^{6x}=3^{x^2}\\\\(3^{\frac{17}{\not6}})^{\not6x}=3^{x^2}[/tex]

[tex]3^{17x}=3^{x^2}\\\\17x=x^2\\\\17x-x^2=0\\\\x(17-x)=0\\\\x=0\ \ \ \ \vee\ \ \ \ 17-x=0\\\\x=0\ \ \ \ \vee\ \ \ \ -x=-17\ \ |\cdot(-1)\\\\x=0\ \ \ \ \vee\ \ \ \ x=17[/tex]

Zastosowano wzory

[tex]\sqrt[n]{a^m}=a^{\frac{m}{n}}\\\\a^m\cdot a^n=a^{m+n}\\\\\frac{a^m}{a^n}=a^{m-n}[/tex]