Question

alguém me ajuda pfv, d)log de (x+1) de base 2 + log de (x+1) de base 4 = 9 sobre 4​

Answer 1

Resposta:

Explicação passo-a-passo:

Vamos mudar o segundo log para a base 2

$log_2(x+1)+log_4(x+1)=\dfrac{9}{4}\\\\log_2(x+1)+\dfrac{log_2(x+1)}{log_24}=\dfrac{9}{4}\\\\log_2(x+1)+\dfrac{log_2(x+1)}{2}=\dfrac{9}{4}\\\\log_2(x+1)+\dfrac{1}{2}log_2(x+1)=\dfrac{9}{4}\\\\log_2(x+1)+log_2(x+1)^{\frac{1}{2}}=\dfrac{9}{4}\\\\log_2(x+1)(x+1)^{\frac{1}{2}}=\dfrac{9}{4}\\\\log_2(x+1)^{\frac{3}{2}}=\dfrac{9}{4}\\\\\dfrac{3}{2}log_2(x+1)=\dfrac{9}{4}\\\\log_2(x+1)=\dfrac{9}{4}.\dfrac{2}{3}\\\\log_2(x+1)=\dfrac{3}{2}\\\\x+1=2^{\frac{3}{2}}\\\\x=\sqrt{2^{3}}-1$

$x=2\sqrt{2}-1$