• Matéria: Matemática
  • Autor: didisouzarj
  • Perguntado 9 anos atrás

me ajudem em uma questao de matemática
Sejam a e b as raízes da equação x²- 100x + 1000 = 0 . Determine o valor de :
Log a + Log b - Log b - Log (a+b)

não consegui fazer esta questão podem me ajudar


didisouzarj: por favor
didisouzarj: correção
Anônimo: Faça a correção!
didisouzarj: log a + log b - log (a+b)
didisouzarj: desculpa
Anônimo: Veja se consegue entender!

Respostas

respondido por: Anônimo
0
Boa noite Didi!

Solução!

 x^{2} -100x+1000=0\\\\\\\
x= \dfrac{100\pm \sqrt{(-100)^{2}-4.1.1000 } }{2} \\\\\\\\\
x= \dfrac{100\pm \sqrt{10000-4000 } }{2} \\\\\\\\\\
x= \dfrac{100\pm \sqrt{6000} }{2} \\\\\\\\\\
x= \dfrac{100\pm 20\sqrt{15} }{2} \\\\\\\\\\
x= \dfrac{2(50\pm 10\sqrt{15} )}{2} \\\\\\\\\\
x= (50\pm 10\sqrt{15} )\\\\\\\\\\

a=50+ \sqrt{15}\\\\\
b=50- \sqrt{15}


loga+logb-log(a+b)\\\\\\\
log(50+10 \sqrt{15}) +log(50-10 \sqrt{15})-log(50+10 \sqrt{15} +50-10 \sqrt{15})\\\\\\\

log[(50+10 \sqrt{15}) \times(50-10 \sqrt{15})]-log(2.(50)\\\\\\\
log(2500-500 \sqrt{15}+500 \sqrt{15}-(10 \sqrt{15})^{2} -log(100)\\\\\\\\\
log(2500-(10 \sqrt{15})^{2})-log(100)\\\\\\\
log(2500-100.15)-log(100)\\\\\\
log(2500-1500)-log(100)\\\\\
log1000-log100


log1000\\\\\\
log1000=x\\\\\\\
10^{x}=1000\\\\\
10^{x}=10^{3}\\\\\\
\boxed{x=3}\\\\\\\\
log100\\\\\\
log100=x\\\\\\
10^{x} =100\\\\\\
10^{x}=10^{2}\\\\\\
\boxed{x=2}


log1000-log100\\\\\
~~~~3~~~~-~~~~2=1

\boxed{Resposta:~~1}

Boa noite!
Bons estudos!

respondido por: genilsonpereiradc73
0

Resposta:

Explicação passo-a-passo:

Perguntas similares