Question

Resolva: log3 (5x^2 – 6x + 16) = log3 (4x^2 + 4x – 5)

Answer 1

Resposta:

S={3,7}

Explicação passo-a-passo:

log₃ (5x² – 6x + 16) = log₃ (4x² + 4x – 5)

5x² – 6x + 16 = 4x² + 4x – 5

x²-10x+21=0

$\displaystyle Aplicando~a~f\'{o}rmula~de~Bhaskara~para~x^{2}-10x+21=0~~\\e~comparando~com~(a)x^{2}+(b)x+(c)=0,~temos~a=1{;}~b=-10~e~c=21\\\\\Delta=(b)^{2}-4(a)(c)=(-10)^{2}-4(1)(21)=100-(84)=16\\\\x^{'}=\frac{-(b)-\sqrt{\Delta}}{2(a)}=\frac{-(-10)-\sqrt{16}}{2(1)}=\frac{10-4}{2}=\frac{6}{2}=3\\\\x^{''}=\frac{-(b)+\sqrt{\Delta}}{2(a)}=\frac{-(-10)+\sqrt{16}}{2(1)}=\frac{10+4}{2}=\frac{14}{2}=7\\\\S=\{3,~7\}$