Question

Resolva as equações logaritimicas propostas:A) log x na base 2x = - 2B) log 4 x na base 2x-3 = 2C) log base 27 log base 2 log (x-1) base 2 4 x = 1/3D) log x na base x-3 = 2

Answer 1

$\mathrm{\mathbf{a)}\ \log_{2x}{x}=-2\ \to\ (2x)^{-2}=x\ \to\ \dfrac{1}{4x^2}=x}\\\\ \mathrm{\dfrac{1}{4}=x^3\ \to\ x=\sqrt[3]{\dfrac{1}{4}}\ \to\ x=\dfrac{1}{\sqrt[3]{4}}\dfrac{\sqrt[3]{2}}{\sqrt[3]{2}}\ \to\ \boxed{\mathrm{x=\dfrac{\sqrt[3]{2}}{2}}}}$

$\mathrm{\mathbf{b)}\ \log_{(2x-3)}{4x}=2\ \to\ (2x-3)^2=4x}\\\\ \mathrm{(2x)^2+2.2x.(-3)+(-3)^2=4x\ \to\ 4x^2-16x+9=0}\\\\ \mathrm{x=\dfrac{-(-16)\pm\sqrt{(-16)^2-4.4.9}}{2.4}=\dfrac{16\pm\sqrt{256-144}}{8}=}\\\\ \mathrm{=\dfrac{16\pm\sqrt{112}}{8}=\dfrac{16\pm4\sqrt{7}}{8}\ \to\ \boxed{\mathrm{x=2\pm\dfrac{\sqrt{7}}{2}}}}$

$\mathrm{\mathbf{c)}\ Ileg\'ivel.}$

$\mathrm{\mathbf{d)}\ \log_{(x-3)}{x}=2\ \to\ (x-3)^2=x}\\\\ \mathrm{x^2+2.x.(-3)+(-3)^2=x\ \to\ x^2-7x+9=0}\\\\ \mathrm{x=\dfrac{-(-7)\pm\sqrt{(-7)^2-4.1.9}}{2.1}=\dfrac{7\pm\sqrt{49-36}}{2}=}\\\\ \mathrm{=\dfrac{7\pm\sqrt{13}}{2}\ \to\ \boxed{\mathrm{x=\dfrac{7}{2}\pm\dfrac{\sqrt{13}}{2}}}}$