Question

ache o valor de x na equação: log x na base 2 = 1/2 + log (4x+10) na base 4

Answer 1

$log_2\ x=\frac{1}{2}+log_4\ (4x+10)\\\\ log_2\ x-log_4\ (4x+10)=\frac{1}{2}\\\\ \frac{log_4\ x}{log_4\ 2}-log_4\ (4x+10)=\frac{1}{2}\\\\ \frac{log_4\ x}{\frac{1}{2}}-log_4\ (4x+10)=\frac{1}{2}\\\\ 2log_4\ x-log_4\ (4x+10)=\frac{1}{2}\\\\ log_4\ x^2-log_4\ (4x+10)=\frac{1}{2}$

$log_4\ \frac{x^2}{4x+10}=\frac{1}{2}\\\\ 4^\frac{1}{2}=\frac{x^2}{4x+10}\\\\ 2(4x+10)=x^2\\\\ x^2=8x+20\\\\ \boxed{x^2-8x-10=0}$

$\Delta=b^2-4(a)(c)\\\\ \Delta = 64+80\\\\ \boxed{\Delta=144}\\\\ x' = \frac{8+12}{2}=\frac{20}{2}=10\\\\ x''=\frac{8-12}{2}=\frac{-4}{2}=-2$

Descartamos a solução negativa, portanto, o valor de x é 10

x=10