Question

resolva as equacoes logx+1 (x elevado a 2 +7) =2

Answer 1

$log_{x+1} x^{2}+7 = 2 \\ \\ \frac{logx^{2}+7}{logx+1}= 2 \\ \\ logx^{2}+7=2.logx+1 \\ \\ logx^{2}+7=log(x+1)^{2} \\ \\ x^{2}+7 = x^{2}+2x + 1 \\ \\ x^{2}-x^{2}-2x=1-7 \\ \\ -2x=-6 \\ \\ x= \frac{6}{2} \\ \\ x=3$

Espero ter ajudado.