Question

logaritmo (10+3x)=2 na base x

Answer 1

$\boxed{\mathrm{\log_b{a}=x \Longleftrightarrow b^x=a}}\\\\ \mathrm{\log_x{(10+3x)}=2\Longleftrightarrow x^2=10+3x}\\ \mathrm{x^2-3x-10=0\ \to\ a=1;\ b=-3;\ c=-10}\\\\ \mathrm{x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}=\dfrac{-(-3)\pm\sqrt{(-3)^2-4.1.(-10)}}{2.1}=}\\\\ \mathrm{=\dfrac{3\pm\sqrt{9+40}}{2}=\dfrac{3\pm\sqrt{49}}{2}=\dfrac{3\pm7}{2}=\boxed{5}\ ou\ \boxed{-2}}\\\\ \textbf{Solu\c{c}\~ao:}\ \boxed{\boxed{\mathrm{x=\{-2;5\}}}}$