Question

Să serezolve ecuația log în baza 5 din (x²+2x-3)=1

Answer 1

 

$\displaystyle\bf\\log_5(x^2+2x-3)=1\\\\5^1=x^2+2x-3\\\\x^2+2x-3=5\\\\x^2+2x-3-5=0\\\\x^2+2x-8=0\\\\x_{12}=\frac{-b\pm\sqrt{b^2-4ac}}{2}\\\\x_{12}=\frac{-2\pm\sqrt{4+32}}{2}\\\\x_{12}=\frac{-2\pm\sqrt{36}}{2}\\\\x_{12}=\frac{-2\pm6}{2}\\\\x_{12}=-1\pm3\\\\x_1=-1-3\\\\x_2=-1+3\\\\\boxed{\bf~x_1=-4}\\\\\boxed{\bf~x_2=2}$