Question

If $x$ is the solution of $\log _4\left(2 x^2\right)+\log _4\left(\frac{1}{8 x}\right)=\log _\pi\left(\tan \frac{\pi}{4}\right)$, then a possible solutions of the equation $y^2+(x+1) y+x=0$, is

• A. $y=1$
• B. $y=-4$
• C. $y=\log _\pi 4$
• D. $y=-1$

Answer 1

Answer: (B), (D)

$\log _4\left(\frac{2 x^2}{8 x}\right)=0 \Rightarrow \log _4\left(\frac{x}{4}\right)=0 \Rightarrow \frac{x}{4}=1 \Rightarrow x=4$
Hence, $y ^2+5 y +4=0 \Rightarrow( y +4)( y +1)=0 \Rightarrow y =-4$ or $y =-1$.